-
I HC THI NGUYN
TRNG I HC KHOA HC
L MINH TIN
CC TR
CA MT S HM NHIU BIN
C CC DNG C BIT
LUN VN THC S TON HC
Thi Nguyn - Nm 2013
-
I HC THI NGUYN
TRNG I HC KHOA HC
L MINH TIN
CC TR
CA MT S HM NHIU BIN
C CC DNG C BIT
Chuyn ngnh: PHNG PHP TON S CP
M s : 60460113
NGI HNG DN KHOA HC
TS. HONG VN HNG
Thi Nguyn - Nm 2013
-
Mc lc
Li ni u 2
1 Mt s cc bt ng thc c in 5
1.1 Bt ng thc Cauchy v cc h qu . . . . . . . . . . . . . . . . .
5
1.2 Bt ng thc Holder v cc h qu . . . . . . . . . . . . . . . . .
9
2 Cc tr ca mt s hm nhiu bin dng c bit 13
2.1 Gi tr b nht ca cc phn thc k- chnh quy . . . . . . . . . . .
13
2.2 Mt s v d p dng nh l 2.1.1 . . . . . . . . . . . . . . . . .
. 16
2.3 Tch ca phn thc k- chnh quy vi phn thc l- chnh quy . . .
21
2.4 Cc tr ca t s hai phn thc ng dng . . . . . . . . . . . . . .
23
2.5 Cc tr ca cc hm na cng tnh . . . . . . . . . . . . . . . . .
. 28
3 Cc tr ca cc hm ca hai a thc i xng hai bin 35
3.1 Cc a thc i xng ca hai bin . . . . . . . . . . . . . . . . .
. 35
3.2 Cc tr ca cc hm ca hai a thc i xng ca hai bin . . . . 36
Kt lun 44
Ti liu tham kho 45
1
-
Li ni u
Cc bi ton cc tr l mt trong nhng vn quan trng ca c ton hc
cao cp ln ton hc s cp. Trong chng trnh ton s cp bc ph thng
trung hc, gi tr b nht hoc ln nht ca cc hm mt bin hoc nhiu
bin
trn mt min no c tm bng mt trong cc phng php sau y:
- Dng o hm kho st hm s trn min cho (i vi hm mt bin).
- Dng l thuyt v tam thc bc hai (phng php min gi tr).
- S dng cc bt ng thc khc nhau.
Cc bi ton tm gi tr b nht hoc ln nht ca hm nhiu bin (theo
thut ng ca ton cao cp) rt thng hay xut hin cu hi phn loi ca
cc thi tuyn sinh i hc v cao ng mn Ton hc (xem thi tuyn
sinh i hc mn Ton cc khi A,B,D t nm 2002 n nm 2012). gii cc
bi ton ny thng phi vn dng phng php th ba (phng php dng
bt ng thc), hai phng php u hu nh khng pht huy tc dng. V
vy, nng cao kh nng gii cc bi ton cc tr ca cc hm nhiu bin,
hc sinh phi rn luyn rt nhiu v cc bt ng thc. Cc bt ng thc m
hc sinh ph thng thng s dng l bt ng thc Cauchy, bt ng thc
Cauchy Schwarz (hay cn gi l bt ng thc Bunhiacovski). Cc bt
ng
thc ny c cc dng tng qut khc nhau, chng hn tng qut hn bt ng
thc Cauchy l bt ng thc Cauchy suy rng, tng qut hn bt ng thc
Cauchy - Schwarz l bt ng thc Holder.
Tc gi Phan Huy Khi (xem [P.H.Khi1]) s dng bt ng thc Cauchy
suy rng tm cc tiu ca mt lp cc hm nhiu bin m tc gi gi l cc
2
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phn thc chnh quy. Theo hng ny, trong [H.V. Hng 1] tc gi Hong
Vn
Hng dng bt ng thc Holder tm gi tr b nht hoc ln nht ca
mt lp cc hm nhiu bin c dng t s ca hai phn thc ng dng. Hng
nghin cu ny cho thy mi mt bt ng thc cha nhiu bin hu nh cho
kh nng kt lun v cc tr ca mt lp no cc hm nhiu bin. Vn l
cc lp hm a ra phi n gin v dng d nhn bit, c nh vy ngi
s dng mi c th nh v vn dng c linh hot.
Bn lun vn Cc tr ca mt s hm nhiu bin c cc dng c
bit nghin cu mt s bt ng thc cha nhiu bin v a ra kt lun tng
ng v gi tr b nht hoc ln nht ca mt s lp hm nhiu bin c dng c
bit. Ni dung ca bn lun vn gm Li ni u, 03 chng v Phn kt lun:
Chng 1: Mt s cc bt ng thc c in
Chng 2: Cc tr ca mt s hm nhiu bin dng c bit
Chng 3: Cc tr ca cc hm ca hai a thc i xng hai bin
Trong chng 1 tc gi a ra cc bt ng thc c in Cauchy, Bunhia-
covski, Holder, Mincowski cng vi mt s h qu v m rng ca chng. Tt
c
cc bt ng thc c a ra u c chng minh cht ch, theo cch ngn
gn nht m tc gi bit.
Chng 2 dnh cho cc nh l v gi tr b nht ca cc phn thc k chnh
quy, gi tr ln nht v b nht ca cc hm c dng t s ca hai phn thc
ng dng, suy rng kt qu cho t s ca cc hm mt bin c dng c bit,
gi tr b nht v ln nht ca cc hm na cng tnh. Sau mi nh l u c
cc v d minh ho ly t cc tuyn sinh i hc mn Ton ca cc khi A,
B, D t nm 2002 n 2012 v mt s ti liu tham kho v cc bi ton cc
tr, bt ng thc. Tt c cc nh l u c chng minh cht ch.
Chng 3 dnh cho vic xt bi ton gi tr b nht v ln nht ca cc hm
hai bin biu din c qua cc a thc i xng ca hai bin u = x + y v v
=
xy. Tc gi a ra v chng minh hai nh l v gi tr b nht v ln nht
3
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ca cc hm dng ny. Mt s v d minh ho cho p dng ca cc nh l ca
chng ny c ly t cc tuyn sinh i hc mn Ton ca cc khi A, B,
D t nm 2002 n 2012. Cc v d khc do tc gi sng tc.
Phn ti liu tham kho gm 07 ti liu.
Tc gi xin by t lng bit n su sc ti thy hng dn TS. Hong Vn
Hng, Vin Khoa hc C bn, Trng i hc Hng hi Vit Nam, thy tn
tnh hng dn tc gi trong sut qu trnh chun b lun vn.
Tc gi cng xin chn thnh cm n cc thy c cng tc ti: Trng i
hc khoa hc, Trng i hc s phm, Khoa cng ngh thng tin- i hc
Thi Nguyn, Trng i hc khoa hc- i hc Quc gia H Ni, Trng i
hc s phm H Ni, Trng i hc Hi Phng, Vin Cng ngh Thng tin,
Vin Ton hc- Vin Khoa hc v Cng ngh Vit Nam rt quan tm v to
iu kin cho tc gi hon thnh chng trnh hc tp bc cao hc trong
sut
thi gian qua.
Hi Phng, ngy 10 thng 5 nm 2013
Tc gi
L Minh Tin
4
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Chng 1
Mt s cc bt ng thc c in
Chng ny chng minh mt s cc bt ng thc c in nh bt ng thc
Cauchy, bt ng thc Cauchy Bunhiacovski, bt ng thc Holder v cc
m
rng ca chng. Cc chng minh a ra trong chng ny khng ging cc
cch
chng minh truyn thng ca cc bt ng thc nu trn trong cc sch ph
thng v ch bt ng thc.
1.1 Bt ng thc Cauchy v cc h qu
Mnh 1.1.1: Vi mi s thc x ta c bt ng thc ex x+1. Du ngthc xy ra
khi v ch khi x = 0.
Chng minh. Xt hm f(x) = exx1 ta c f (x) = ex1, f (x) = 0 x =
0.Khi x < 0 ta c f (x) < 0, khi x > 0 ta c f (x) > 0.
Vy f(x) t cc tiu thc
s v ton cc ti x = 0. Ta c f(0) = 0. Vy f(x) 0 vi mi s thc x,
dung thc xy ra khi v ch khi x=0. Khng nh ny tng ng vi bt ng
thc ex x+ 1, du ng thc xy ra khi v ch khi x = 0.nh l 1.1.2 (bt
ng thc Cauchy): Nu xi ( i =1,2,..,n) l n s dng
th:
1
n
ni=1
xi n n
i=1
xi
du ng thc xy ra khi v ch khi x1 = x2 = ... = xn
5
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Chng minh. t A =1
n
ni=1
xi, G = n
ni=1
xi. p dng khng nh ca mnh
1.1.1 vi x =xiA 1 ( i = 1,..,n) ta c: exiA1 xi
A(du ng thc xy ra khi
v ch khixiA
= 1 xi = A).V cc xi u l cc s dng nn t bt ng thc trn ta suy
ra:
ni=1
exiA1
ni=1
(xiA) =
Gn
An e
ni=1
xiAn G
n
An 1 = enn G
n
An
An Gn A G 1n
ni=1
xi n
ni=1
xi
Du ng thc xy ra khi v ch khi xi =A vi mi i tt c cc xi phibng
nhau.
Tip theo ta k hiu Rn+ l tp hp cc vc t x = (x1, ..., xn) ca Rn
m
mi thnh phn xi u dng, Qn+ l tp con ca R
n+ m mi thnh phn ca
cc vc t thuc n u l cc s hu t dng. Ta s gi mi vc t thuc Qn+
l mt vc t hu t dng n- chiu. Hm thc n bin f(x)= f (x1, ..., xn )
gi l
lin tc ti im x* = (x1, ..., xn) Rn nu x* thuc tp xc nh ca f(x)
v
vi mi dy {k = (1k, ..., nk)} nm trong tp xc nh ca f(x) tho
mn:
limk
ik = xi (i {1, ..., n}) (1.1)
ta lun c limk
f(k) = f(x). Khi (1.1) ng ta ni dy {k = (1k, ..., nk)} hi
t ti x* v k hiu limk
k = x.
Nu D l mt tp con ca tp xc nh ca f(x) v f(x) lin tc ti mi x Dta
ni f(x) lin tc trn D. Mt vc t p = (p1, ..., pm) Rm+ gi l mt h
trnglng chun nu
mi=1
pi = 1. Ta c mnh sau:
Mnh 1.1.3: Gi s F(p,x), G(p,x) l hai hm xc nh trn Rm+ Rn+v lin
tc theo bin p trn Rm+ vi mi x c nh Rn+ . Khi :
1) Nu bt ng thc F(p,x) G(p,x) ng vi mi (p,x) Qm+ Rn+ thn cng ng
vi mi (p,x) Rm+ Rn+ .
6
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2) Nu bt ng thc F(p,x) G(p,x) ng vi mi cp (p,x) Qm+ Rn+trong p l
mt h trng lng chun th n cng ng vi mi cp (p,x) Rm+ Rn+ trong p l mt
h trng lng chun.
Chng minh. 1) Gi s (p*,x*) l mt cp tu thuc Rm+ Rn+ . Tnti mt dy
{pk} cc vc t hu t dng sao cho lim
kpk = p
. Khi theo gi
thit ta c:
F(pk,x*) G(pk,x*) vi mi k nguyn dng. (1.2)
Cho k dn ti v cc trong (1.2) v dng tnh lin tc ca cc hm F, G
theo
bin p ta c iu cn chng minh: F(p*,x*) G(p*,x*).2) Chng minh hon
ton tng t nh phn 1). Ch cn nhn xt rng nu
p* l mt h trng lng chun tu th bao gi cng tn ti mt dy cc h
trng lng chun gm ton cc vc t hu t dng {pk} sao cho limk
pk = p.
nh l 1.1.4 (bt ng thc Cauchy suy rng):
Gi s x= (x1, x2...xn) l mt vc t thuc Rn+ v p = (p1, . . . , pn)
l mt h
trng lng chun. Khi ta c bt ng thc:ni=1
pixi ni=1
xpii (1.3)
Chng minh. Trc ht ta chng minh bt ng thc (1.3) ng vi mi
h trng lng chun hu t. Gi s pi l cc s hu t dng sao choni=1
pi = 1.
Quy ng mu s chung cho tt c cc pi ta c th xem pi =mim
trong
mi(i = 1, 2, . . . , n) v m l cc s nguyn dng tho mnni=1
mi = m. p dng
bt ng thc Cauchy cho m s dng: m1 s x1, m2 s x2 ,. . . , mn s xn
ta
c:
1
m
ni=1
mixi m
ni=1
xmii ni=1
mimxi
ni=1
xmi/mi
ni=1
pixi ni=1
xpii
Vy (1.3) ng vi mi cp (p,x) Qn+ Rn+ . t F(p,x) =ni=1
pixi, G(p,x)
=ni=1
xpii
7
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R rng F(p,x) v G(p,x) l cc hm lin tc theo p = (p1, . . . , pn)
trn min
Rn+ vi mi x c nh thuc Rn+ . p dng khng nh 2) ca mnh 1.1.3 ta
suy ra tnh ng n ca bt ng thc Cauchy suy rng (1.3).
Mnh 1.1.5: Nu xj(j = 1, 2, ..., k) l cc s dng tho mnkj=1
xj = 1
v / > 1 th:kj=1
xj kj=1
xj (1.4)
Nhn xt: Bt ng thc (1.4) cho cu tr li khng nh i vi cu hi
t ra bi tc gi L Thng Nht trong bi bo Nhng suy ngh ban u v
thi tuyn sinh i hc mn Ton nm 2001 (Tp ch Ton hc v Tui tr s
8/2001)
Chng minh. Chng minh bt ng thc (1.4) c chia lm hai bc:
i) Bc 1: Ta s chng minh rng nu m, n l cc s nguyn dng v
m>n
th:kj=1
xm/nj
kj=1
xj (1.5)
Vi mi j {1, 2, ..., k} p dng bt ng thc Cauchy cho m s dng: n
sbng xm/nj v m n s bng 1 ta c:
nxm/nj +m n mxj (j = 1, 2, ..., k) (1.6)
t A =kj=1
xm/nj , cng cc v ca k bt ng thc trong (1.6) ri p dng
bt ng thc Cauchy cho k s dng xj(j = 1, 2, ..., k) (ch kj=1
xj = 1) ta suy
ra:
nA+ k(m n) mkj=1
xj = nkj=1
xj + (m n)kj=1
xj
nkj=1
xj + k(m n) A kj=1
xj
Vy bt ng thc (1.5) ng, tc bt ng thc (1.4) ng vi l s hu
t =m
n, m, n l cc s nguyn dng, m > n v = 1
8
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ii) Bc 2: Gi s > 1 l s v t, chn dy s hu t (rn) sao cho rn
> 1
v limn rn = . Bi v (1.5) ng khi =
m
nl s hu t > 1 ta c:
kj=1
xrnj kj=1
xj (vi mi n )
p dng mnh 1.1.3 vi p = R+, x = (x1, ..., xk) Rk+ , F(p,x) =
F(,x)=
ki=1
xi , G(p,x) = G(,x) =kj=1
xj (G khng ph thuc vo ) do F, G u lin
tc theo ta c:kj=1
xj kj=1
xj (1.7)
Kt hp cc kt qu ca bc 1 v bc 2 ta suy ra bt ng thc (1.7) ng
vi mi s thc > 1. Thay trong (1.7) xj bi xj v bi
> 1 ta thu c
(1.4):kj=1
xj kj=1
xj
1.2 Bt ng thc Holder v cc h qu
Mnh 1.2.1: Gi s k, k l cc s dng tho mn1
k+
1
k= 1. Khi
vi mi x, y dng ta c bt ng thc:
xy xk
k+yk
k(1.8)
Du ng thc xy ra khi v ch khi xk = yk
Chng minh. C nh y v xt hm f(x) = xy xk
k y
k
k. o hm theo
bin x ta c :
f (x) = y xk1; f (x) = 0 xk = yk
Xt du o hm f(x) trn min x > 0 ta c f(x) > 0 khi x yk1. Vy
hm f(x) t cc i cht v ton cc trn min
x>0 ti x = yk1. Ta c f
(yk
1) = 0. Vy f (x) 0 vi mi x > 0, du ng9
-
thc ch xy ra khi v ch khi x = yk1 hay xk = yk
. Khng nh ny tng
ng vi khng nh ca mnh 1.2.1 v (1.8) c chng minh.
Nhn xt: i) Bt ng thc (1.8) vn ng nu mt trong cc s x, y hoc
c hai u bng 0. Nu x = 0, y > 0 hoc x >0, y = 0 th trong
(1.8) c du
bt ng thc thc s.
ii) Bt ng thc (1.8) c th suy ra t bt ng thc Cauchy suy rng
(1.3).
Thc vy, t trong (1.3) n = 2, x1 = xk, x2 = yk. V
1
k+
1
k= 1 p dng (1.3)
vi p1 =1
k, p2 =
1
kta c:
xp11 xp22 p1x1 + p2x2 xy
xk
k+yk
k
Mnh 1.2.2 (bt ng thc Holder): Gi s a = (a1, ..., an),
b = (b1, ..., bn) l hai vc t ca Rn, k v k l hai s dng tho mn
1
k+
1
k= 1.
Khi ta c:ni=1
|aibi| (ni=1
|ai|k)1/k(ni=1
|bi|k)1/k
(1.9)
Du ng thc xy ra khi v ch khi cc vc t (|a1|k, ..., |an|k),
(|b1|k, ..., |bn|k
)
ph thuc tuyn tnh.
Chng minh. R rng nu mt trong hai vc t a hoc b l vc t khng
th (1.9) hin nhin ng. Do ta ch cn chng minh (1.9) vi gi thit l
v
phi ca (1.9) dng. Khng gim tng qut ta c th xem trong (1.9) tt
c
cc ai , bi u khng m ng thi t nht mt ai v mt bj no dng. Thay
ln lt trong (1.8) x v y bi :
xi =ai
(ni=1
aki )1/k
, yi =bi
(ni=1
bki )
1/k(i = 1, ..., n)
ta c :
xiyi xik
k+yik
k
aibi(ni=1
aki )1/k
(ni=1
bki )
1/k a
ki
k(ni=1
aki )
+bk
i
k(ni=1
bki )
(i = 1, ..., n) (1.10)
10
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Ly tng theo i cc bt ng thc (1.10) ta c:
ni=1
xiyi
ni=1
xki
k+
ni=1
yki
k=
1
k+
1
k= 1
ni=1
aibi (ni=1
aki )1/k(
ni=1
bki )
1/k (1.11)
Du ng thc trong (1.11) xy ra khi v ch khi c du ng thc tt c
cc
bt ng thc (1.10), tc l hoc ai = bi = 0 hocakini=1
aki
=bk
i
ni=1
bki
aik = bik
vi =
ni=1
aki
ni=1
bki
cc vc t (ak1, . . . , akn), (bk1, .., bkn), ph thuc tuyn
tnh.
Trng hp ring ca bt ng thc Holder l bt ng thc Cauchy- Bun-
hiacovski.
Mnh 1.2.3 (bt ng thc Cauchy-Bunhiacovski): Gi s
a = (a1, ..., an), b = (b1, ..., bn) l hai vc t ca Rn. Khi ta
c:
ni=1
|aibi|
( ni=1
a2i )(
ni=1
b2i ) (1.12)
Du ng thc xy ra khi v ch khi cc vc t (|a1| , ..., |an|), (|b1| ,
..., |bn|) phthuc tuyn tnh.
Chng minh. p dng bt ng thc Holder vi k = k = 2 ta suy ra bt
ng thc (1.12). Du ng thc xy ra khi v ch khi cc vc t (|a1|2, ...,
|an|2),(|b1|2, ..., |bn|2) ph thuc tuyn tnh cc vc t (|a1| , ...,
|an|), (|b1| , ..., |bn|) phthuc tuyn tnh.
Mnh 1.2.4 (bt ng thc Mincowski): Vi s k 1 v a =(a1, ..., an), b
= (b1, ..., bn) l hai vc t ca R
n ta c:(ni=1
|ai + bi|k)1/k
(
ni=1
|ai|k)1/k
+
(ni=1
|bi|k)1/k
(1.13)
11
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Chng minh. R rng (1.13) ng nu v tri ca n bng 0. Cng hin
nhin l (1.13) ng vi k = 1 do bt ng thc | + | ||+ || ng vi mis
thc , . Vy ta ch cn chng minh (1.13) ng khi v tri ca n dng
v k > 1. Gi k l s tho mn1
k+
1
k= 1. p dng bt ng thc (1.9) ta c:
ni=1
|ai + bi|k =ni=1
|ai + bi| |ai + bi|k1 ni=1
|ai| |ai + bi|k1 +ni=1
|bi||ai + bi|k1
(ni=1
|ai|k)1/k(ni=1
|ai + bi|k(k1))1/k
+ (
ni=1
|bi|k)1/k(ni=1
|ai + bi|k(k1))1/k
= (
ni=1
|ai|k)1/k(ni=1
|ai + bi|k)1/k+ (
ni=1
|bi|k)1/k(ni=1
|ai + bi|k)1/k
(1.14)
Dng gi thit v tri ca (1.13) dng, t (1.14) ta suy ra:
(ni=1
|ai + bi|k)1 1k (ni=1
|ai|k)1/k + (ni=1
|bi|k)1/k
(ni=1
|ai + bi|k) 1k (ni=1
|ai|k)1/k + (ni=1
|bi|k)1/k
Vy (1.13) c chng minh.
12
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Chng 2
Cc tr ca mt s hm nhiu bindng c bit
Chng ny s dng cc bt ng thc chng minh trong chng 1
a ra cc kt lun v gi tr b nht hoc ln nht ca mt s hm nhiu bin
c dng c bit.
2.1 Gi tr b nht ca cc phn thc k- chnh quy
nh ngha 2.1.1. Cho k l mt s thc. Hm ca n bin x1, ..., xn
dng:
f =mi=1
cix1i1 x
2i2 ...x
nin (2.1)
trong ci l cc s dng, ji ( i = 1,...,m; j = 1,...,n ) l cc s thc
ty ,
x = (x1, ..., xn) Rn+ gi l mt phn thc k chnh quy nu vi mi j =
1,...,nta u c:
mi=1
ciji = k (2.2)
Nhn xt : Khi k = 0, tc gi Phan Huy Khi (xem [P.H.Khi 1]) gi
mt
hm n bin dng (2.1) l mt phn thc chnh quy.
nh l 2.1.1: Gi s f =mi=1
cix1i1 x
2i2 ...x
nin l mt phn thc k chnh quy.
t:
t = x1x2...xn
khi :
13
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1. Vi k = 0 gi tr b nht ca f trn Rn+ l S =mi=1
ci (xem [P.H.Khi 1]).
2. Vi k > 0 gi tr b nht ca f trn min {t 1} l S =mi=1
ci
3. Vi k < 0 gi tr b nht ca f trn min {0 < t 1} l S
=mi=1
ci
4. Vi k tu , gi tr b nht ca f trn min t = 1 l S =mi=1
ci
Chng minh.
1) Gi s x = (x1, ..., xn) Rn+ t:
pi =ciS, yi = x
1i1 ...x
nin (i = 1, ...,m)
ta c p = (p1, p2, ..., pm) l mt h trng lng chun v y = (y1, y2,
..., ym) Rm+ .p dng bt ng thc Cauchy suy rng (1.3) cho h trng lng
chun p v
vc t y Rm+ , dng iu kin (2.2) vi k = 0 ta c:mi=1
piyi mi=1
ypii = 1 f =mi=1
cix1i1 ...x
nin S =
mi=1
ci (trn min x Rn+ ).
Mt khc f(1, 1, ..., 1) =mi=1
ci = S. Vy:
Min{f : x = (x1, ..., xn) Rn+ } = S.
2) Gi s k > 0, t g = f+xk1 ...xkn = f+ t
k. Khi g l mt phn thc chnh
quy ca n bin x1, ..., xn. Thc vy, t cm+1 = 1, j (m+1) = k (j =
1, ..., n) iukin (2.2) i vi g tr thnh:
m+1i=1
ciji = 0 (j = 1, ..., n)
Theo chng minh khng nh 1) ta c g(x1, ..., xn) S =m+1i=1
ci =mi=1
ci + 1.
Do trn min {t 1} ta c f(x1, ..., xn) S tk S 1 = S =mi=1
ci (ch
rng trn min t > 0 hm tk nghch bin). Vi x1 = ... = xn = 1 ta c
t = 1 v
f(1,. . . ,1) = S. Vy suy ra :
Min {f : t 1} = S
14
-
3) Gi s k < 0. L lun tng t nh trong chng minh phn 2) ta
c:
f(x1, ..., xn) S tk S 1 = S =mi=1
ci
(ch rng trn min {0 < t 1} hm tk ng bin). Vi x1 = . . .= xn =
1 tac t = 1 v f(1,. . . ,1) = S. Vy suy ra:
Min {f : 0 < t 1} = S
4) By gi gi s k l s thc tu v t = 1. L lun hon ton tng t
nh trong chng minh phn 2) v phn 3) ta c :
f(x1, ..., xn) S tk = S 1 = S =mi=1
ci
Li do vi x1 = . . .= xn = 1 ta c t = 1 v f(1,. . . ,1) = S. Vy
suy ra:
Min {f : t = 1} = S
nh l 2.1.1 c chng minh hon ton.
Nhn xt: i) Khi k 6= 0 mt phn thc k chnh quy c th khng c gitr b
nht v ln nht trn min Rn+ nh cc v d sau y chng t:
a) Hm f(x, y) =1
x+
1
yl phn thc (-1)- chnh quy ca 2 bin x,y. Ta c
f(x,x)=2
xnn d dng suy ra sup
{f : (x, y) R2+
}=, inf{f : (x, y) R2+ } =0.
Nhng f(x,y)> 0 vi mi x,y R2+b) Hm g(x, y, z) =
xy
z+yz
x+zx
yl phn thc 1- chnh quy ca 3 bin x,y,z.
Ta c g(x,x,x) = 3x nn cng d dng suy ra:
sup{g : (x, y, z) R3+
}= +, inf {g : (x, y, z) R3+ } = 0
Nhng g(x,y,z)> 0 vi mi (x,y,z) R3+ii) Trn min t = 1, vi mi s
thc k lun tn ti cc phn thc k chnh
quy khng c gi tr ln nht trn min ny. Chng hn, nu k = 0, hm:
h(x, y) = x2 +1
x2+ y2 +
1
y2
15
-
l phn thc chnh quy ca 2 bin x, y khng c gi tr ln nht trn min
t=xy=1. Bi v trn min ny tp gi tr ca h trng vi tp gi tr ca hm
mt bin:
h(x, x) = 2(x2 +1
x2) (x > 0)
R rng sup{h (x, x) : x > 0}= +.Nu k 6= 0, hm:
p(x, y) = xk + yk
l phn thc k chnh quy tho mn sup{p (x, y) : t = xy = 1} = +iii)
Trn min {t = 1}, mt phn thc k chnh quy ca n bin c th bin
i thnh mt phn thc chnh quy ca n 1 bin trn R(n1)+ . V vy kt
lun
ca khng nh 4) trong nh l 2.1.1 cng c th suy ra t khng nh 1)
ca
nh l v nhn xt ny.
2.2 Mt s v d p dng nh l 2.1.1
p dng nh l 2.1.1 c th tm gi tr b nht ca mt s cc hm nhiu
bin trn mt tp con ca min m cc bin ch nhn gi tr dng. Trc
ht ta ch ra mt vi v d p dng trc tip cc khng nh ca nh l
2.1.1,
sau cc v d phc tp hn, xut hin trong cc ti liu khc v cc bi
ton
cc tr hoc bt ng thc s c ch ra.
V d 1: Tm gi tr b nht ca hm s sau trn min m cc bin ch nhn
gi tr dng:
f(x, y, z) = xyz +
x+
y+
z( , , l cc hng s dng)
Gii. Hm f l phn thc chnh quy ca 3 bin x,y,z bi v:
1. + (1) = 1. + .(1) = 1. + .(1) = 0
Theo khng nh 1) ca nh l 2.1.1 ta c min{f : (x, y, z) R3+
}= 1++
+
16
-
V d 2: Tm gi tr b nht ca hm g di y trn min cc bin x, y, z
nhn gi tr dng v tho mn xyz 1:
g(x, y, z) =x2+2.y2+2
z22+
2
y+z2
x+
1
x
( l hng s dng, l hng s thc tu ).
Gii. Hm g l phn thc 2 chnh quy, bi v :
1.(2 + 2) + .(1) + 1.() = 21. (2 + 2) + 2. () = 21. (2 + 2) + .2
= 2
Theo khng nh 2) ca nh l 2.1.1 gi tr b nht ca hm g trn min
c ch ra l:
1+2+ +1 = 4+
V d 3: Tm gi tr b nht ca hm h cho di y trn min
D= {x > 0, y > 0, z > 0;x+ y + z 3}
h(x, y, z) =1
x2+
1
y2+
1
z2+x3 + y3 + z3 xyz
2
Gii. p dng bt ng thc Cauchy vi (x,y,z) thuc min D ta c:
xyz (x+ y + z3
)3 1 (2.3)
x3 + y3 + z3 3xyz
Do : h(x, y, z) 1x2
+1
y2+
1
z2+3xyz xyz
2=
1
x2+
1
y2+
1
z2+xyz = N(x, y, z)
Hm N(x,y,z) l phn thc (-1) chnh quy. Theo khng nh 3) ca nh l
2.1.1, trn min G= = {x > 0, y > 0, z > 0;xyz 1} ta c
N(x,y,z) 1 + 1 + 1 +1 = 4. Do trn min G ta cng c h(x,y,z) 4. Bt ng
thc (2.3) chngt min D nm trong min G. Vy trn D ta c:
h(x,y,z) 4
17
-
Nhng im (1,1,1) thuc min D v h(1,1,1) = 4 do ta c:
min{h (x, y, z) : (x, y, z) D} = 4
V d 4: Tm gi tr b nht ca hm k cho di y trn min
E = {x > 0, y > 0, z > 0;xyz = 1}:
k(x, y, z) = x3 +5
x+
2
y+
4z
Gii. D thy hm k(x,y,z) l hm (-2) chnh quy, do theo khng nh
4) ca nh l 2.1.1 ta c ngay min{k (x, y, z) : (x, y, z) E} = 1 +
5 + 2 + 4 = 12.Bn c khng kh khn c th t to ra cc v d vi cc hm phc
tp
hn. By gi ta s chng minh mt s bt ng thc nh s dng nh l 2.1.1.
V d 5: Cho m, n l cc s nguyn dng v a 0, b 0. Chng minh btng
thc:
m ma+ n
nb (m+ n) m+n
ab (2.4)
(bi ton 1.9 [Tr.Phng ]).
Gii. Bt ng thc (2.4) r rng ng nu ab = 0, do ta ch cn chng
minh (2.4) khi a > 0, b >0. Xt hm:
f(a, b) = m.a
1
m
1
m+ n .b
1m+ n + n.b
1
n
1
m+ n .a
1m+ n
f(a,b) l phn thc chnh quy i vi hai bin a, b v iu kin (2.2) i vi
cc
bin a, b tng ng l:
m(1
m 1m+ n
) + n.(1
m+ n) = 0
m.(1
m+ n) + n.(
1
n 1m+ n
) = 0
Theo khng nh 1) ca nh l 2.1.1 ta suy ra :
f(a, b) m+ n (2.5)
18
-
Nhng (2.5) tng ng vi (2.4) do (2.4) c chng minh.
V d 6: Cho a, b, c l cc s dng. Chng minh bt ng thc :
bc
a+ca
b+ab
c a+ b+ c
(bi ton 5.1 [Tr.Phng]).
Gii. Xt hm s: f(x, y, z) = bx+ cy + az +b
x+c
y+a
z
D thy f l phn thc chnh quy ca 3 bin x,y,z nn theo khng nh 1)
ca nh l 2.1.1 trn min x, y, z dng ta c:
f(x, y, z) b+ c+ a+ b+ c+ a = 2(a+ b+ c)
Thay x =c
a, y =
a
b, z =
b
ctrong biu thc ca f ta c:
f(c
a,a
b,b
c) =
bc
a+ca
b+ab
c+ba
c+cb
a+ac
b= 2(
bc
a+ca
b+ab
c) 2(a+ b+ c)
bca+ca
b+ab
c a+ b+ c (iu phi chng minh)
V d 7: Chng minh rng:
a
mb+ nc+
b
mc+ na+
c
ma+ nb 3m+ n
(2.6)
vi mi a, b, c, m, n dng (bi ton 5.8 [Tr.Phng]).
Gii. t x = mb+ nc, y = mc+ na, z = ma+ nb, ta c x, y, z dng
v:
a =m2z + n2y mnx
m3 + n3; b =
n2z +m2xmnym3 + n3
; c =m2y + n2xmnz
m3 + n3
Thay biu din ca a, b, c qua x, y, z vo v tri ca (2.6) ta c:
1
m3 + n3(m2z
x+n2y
x+n2z
y+m2x
y+m2y
z+n2x
z) 3mn
m3 + n3
t: f(x, y, z) =1
m3 + n3(m2z
x+n2y
x+n2z
y+m2x
y+m2y
z+n2x
z)
D thy f(x,y,z) l phn thc chnh quy, do theo khng nh 1) ca nh
l 2.1.1 ta c:
f(x, y, z) 3(m2 + n2)
m3 + n3
19
-
Vy v tri ca (2.6) 3(m2 + n2)
m3 + n3 3mnm3 + n3
=3
m+ n(iu phi chng minh)
V d 8: Cho n s dng x1, ..., xn tho mn : x1 + x2 + ... + xn S.
Chngminh rng:
ni=1
(1 +1
xi) (1+n
S)n (2.7)
((2.7) l tng qut ho ca b trong li gii bi ton 295 trong [P.H.Khi
2]).
Gii. thun tin cho vic trnh by ta a vo k hiu sau:
Nu a1, ..., an l n bin th ta k hiu
n,k(a) l tngSk
jSk
aj, trong Sk
l mt tp con gm k phn t (k 1) ca tp hp {1, 2, ..., n}. TngSk
c ly
theo tt c cc tp con Sk gm k phn t ca tp hp {1, 2, ..., n} (c Ckn
cc tpSk nh vy). V d, nu n =3 th:
3,1(a) = a1 + a2 + a3,
3,2(a) = a1a2 + a2a3 + a3a1,
3,3(a) = a1a2a3
Vi dy bin dng y1, ..., yn k hiu1
ytng ng vi dy
1
y1, ...,
1
ynPhn thc:
fk =
n,k(1
y)
l phn thc (Ck1n1)- chnh quy. Do , theo khng nh 3) ca nh l
2.1.1nu xt trn min t = y1y2...yn 1 ta c fk Ckn.
Gi s dy s dng x1, ..., xn tho mn x1+x2+ ...+xn S t yi = nxiS
ta c
yi > 0 v y1 + y2 + ...+ yn n. T bt ng thc Cauchy suy ra t =
y1y2...yn 1.Vy:
n,k(1
x) = (
n
S)k
n,k(1
y) Ckn(
n
S)k
Suy ra:
ni=1
(1 +1
xi) = 1 +
nk=1
n.k(
1
x) 1 +
nk=1
Ckn(n
S)k = (1 +
n
S)n
Nh vy (2.7) c chng minh. Du ng thc trong (2.7) xy ra khi v
ch
khi tt c cc xi u bng nhau v bngS
n.
20
-
Nhn xt (bi ton 295 trong [P.H.Khi 2]): Khi A,B,C l cc gc
trong
ca mt tam gic v x1 = sinA, x2 = sinB, x3 = sinC ta c xi > 0
v:
sin A +sin B + sinC 33
2= S
Do p dng (2.7) vi n = 3 ta c:
(1 +1
sinA)(1 +
1
sinB)(1 +
1
sinC) (1 + 2
3)3
2.3 Tch ca phn thc k- chnh quy vi phn thc
l- chnh quy
nh l 2.3.1: Gi s:
f =mi=1
cix1i1 x
2i2 ...x
nin l phn thc k chnh quy.
g =s
p=1
dpx1p1 x
2p2 ...x
npn l phn thc l chnh quy.
Khi hm h(x1, ..., xn) = f(x1, ..., xn)g(x1, ..., xn) l phn thc r
chnh quy,
trong r tnh theo cng thc:
r = k
sp=1
dp + l
mi=1
ci
Ni ring, tch ca hai phn thc chnh quy li l mt phn thc chnh
quy.
Chng minh. Thc hin php nhn phn thc f vi phn thc g v lp tng
dng (2.2) i vi hm h = f.g cho bin xj tu ta c tng sau:
r =
mi=1
sp=1
cidp(ji + jp) =
mi=1
sp=1
cidpji +
mi=1
sp=1
cidpjp
=
sp=1
dp
mi=1
ciji +
mi=1
ci
sp=1
dpjp = k
sp=1
dp + l
mi=1
ci
Nh vy r khng ph thuc j, do h l phn thc r chnh quy.
21
-
H qu 2.3.2: Gi thit nh trong nh l 2.3.1. Khi :
1) Nu r = ks
p=1
dp + lmi=1
ci = 0 th gi tr b nht ca hm h = f.g trn Rn+
l (mi=1
ci)(s
p=1
dp).
2) Nu r = ks
p=1
dp+ lmi=1
ci > 0 th gi tr b nht ca hm h = f.g trn min
{t = x1x2...xn 1} l (mi=1
ci)(s
p=1
dp)
3) Nu r = ks
p=1
dp+ lmi=1
ci < 0 th gi tr b nht ca hm h = f.g trn min
{t = x1x2...xn 1} l (mi=1
ci)(s
p=1
dp).
V d: Tm gi tr b nht trn min D ={x > 0, y > 0, z > 0 :
x+ y + z 3}ca hm s sau:
f(x, y, z) = (x2013y2014z2015 +2012
x+
2013
y+
2014
z)(
1
x2+
1
y2+xy
z)
Gii: t:
g(x, y, z) = x2013y2014z2015 +2012
x+
2013
y+
2014
z; h(x, y, z) =
1
x2+
1
y2+xy
z
D thy g l phn thc 1 chnh quy cn h l phn thc (-1) chnh quy,
do
f l phn thc r chnh quy vi r = (1+2012+2013+2014)(-1) +
(1+1+1).1
= - 6037 < 0. T iu kin x > 0, y > 0, z > 0 v x + y +
z 3, dng btng thc Cauchy ta suy ra 0 < t = xyz 1. Do D c cha
trong min E= {x > 0, y > 0, z > 0; t = xyz 1}. Theo h qu
2.3.2 ta c f(x,y,z) c gi tr bnht trn min E l m = (
1+2012+2013+2014)(1+1+1) = 18120. Do trn
min D ta phi c: f(x,y,z) 18120Nhng im (1,1,1) thuc D v f(1,1,1)
= 18120 do :
min{f (x, y, z) : (x, y, z) D} =18120.
22
-
2.4 Cc tr ca t s hai phn thc ng dng
nh ngha 2.4.1: Xt hm n bin cho bi cng thc (2.1) :
f(x1, ..., xn) =
mi=1
cix1i1 x
2i2 ...x
nin
Cc gi thit v hm f ging nh trong nh ngha 2.1.1, ngoi tr iu
kin
k- chnh quy (2.2). Khi ta ni rng cc hm:
u =f(x1, ..., xn)
k
f(xk1, ..., xkn)
v v =f(xk1, ..., x
kn)
f(x1, ..., xn)k
trong k l s thc tu , k > 1, l cc hm c dng t s ca hai phn
thc
ng dng.
T bt ng thc Holder ta suy ra bt ng thc sau y:
Mnh 2.4.2: Gi s q = (q1, q2, ..., qn), z = (z1, z2, ..., zn) l
cc vc t
thuc Rn+ , k > 1, khi c bt ng thc:
ni=1
qizi (ni=1
qi)1
1
k (
ni=1
qizki )
1
k (2.8)
(xem [Hardy, Littllewood,Polya], trang 85).
Nhn xt : Hin nhin (2.8) vn cn ng nu tng hu hn trong (2.8)
c thay bng chui v hn v mt s cc qi hoc zi bng 0, trong ta xem
(2.8) l ng nu v phi (2.8) bng +. Ta s ch s dng bt ng thc (2.8)di
dng chui km theo gi thit l khng phi tt c cc qi v zi u bng 0.
Chng minh. t ai = q1/k
i , bi = q1/ki .zi ( i =1,. . . ,n), trong
1
k+
1
k= 1.
p dng bt ng thc Holder cho hai dy s (ai) v (bi) ta c ngay
(2.8).
nh l 2.4.3 ([H.V.Hng 1]): t:
f(x1, ..., xn) =
mi=1
cix1i1 x
2i2 ...x
nin
23
-
trong : c = (c1, c2, ..., cm) Rm+ x = (x1, ..., xn) Rn+ 1i, 2i,
..., ni (i =1, 2, ...,m) l cc s thc tu , k > 1. Khi ta c:
max{f(x1, ..., xn)k
f(xk1, ..., xkn)
: x = (x1, ..., xn) Rn+ } = (mi=1
ci)k1
Chng minh. t qi = ci, zi = x1i1 ...x
nin , (i = 1, 2, ...,m). p dng bt ng
thc (2.8) ta c:
f(x1, ..., xn) =
mi=1
cix1i1 x
2i2 ...x
nin (
mi=1
ci)1
1
k (
mi=1
cixk1i1 x
k2i2 ...x
knin )
1
k
f(x1, ..., xn)f(x1k, ..., xnk)
1/k (
mi=1
ci)1
1
k f(x1, ..., xn)k
f(x1k, ..., xnk) (
mi=1
ci)k1
Mt khc, ta c:
f(1, 1, ..., 1)k
f(1k, 1k, ..., 1k)=f(1, 1, ..., 1)k
f(1, 1, ..., 1)= (
mi=1
ci)k1
Vy:
max{f(x1, ..., xn)k
f(xk1, ..., xkn)
: x = (x1, ..., xn) Rn+ } = (mi=1
ci)k1
V d: Tm gi tr ln nht ca cc hm sau trn min m tt c cc bin
u dng:
1) y =(x+ 1)2013
x2013 + 1( x l bin )
2) z =(ax3 + bx2y + cxy2 + dy4)
2
ax6 + bx4y2 + cx2y4 + dy8(a, b, c, d (0;+)) (x,y l bin)
3) u =(x3z + 2y2z3 + 5xz6)
10
x30z5 + 1024y20z30 + 5x10z60(x,y,z l cc bin).
Gii. Xem cc n thc khng cha bin no nh cc n thc cha bin
vi s m 0, ta thy tt c cc gi thit ca nh l 2.4.3 u c tho mn
i vi v d 1 v 2 (cc gi tr tng ng ca k l: k =2013 trong v d 1;
k=2
trong v d 2). i vi v d 3, t t = 2y2 ta vit li hm u nh sau:
u =(x3z + tz3 + 5xz6)
10
x30z5 + t10z30 + 5x10z60
24
-
gi tr k tng ng l 10. Vy theo nh l 2.4.3, gi tr ln nht ca cc
hm
s tng ng trn min tt c cc bin u dng l: max{y : x > 0} =
22012,max{z : x > 0, y > 0}= a+b+c+d, max{u : x > 0, y
> 0, z > 0} = 79
Nhn xt : Bng cch i bin bn c c th tm gi tr ln nht ca cc
hm phc tp hn cc hm a ra trong cc v d trn.
T nh l 2.4.3 ta suy ra h qu:
nh l 2.4.4 [H.V.Hng 1]: Vi cc gi thit nh trong nh l 2.4.3 ta
c:
min{ f(xk1, ..., x
kn)
f(x1, ..., xn)k: x = (x1, ..., xn) Rn+ } = (
mi=1
ci)1k
Chng minh. Vi cc gi thit nh trong nh l 2.4.3, t:
u =f(x1, ..., xn)
k
f(xk1, ..., xkn)
, v =f(xk1, ..., x
kn)
f(x1, ..., xn)k
Khi u > 0, v > 0 trn min Rn+ v v =1
u. Do trn min Rn+ ta c:
min{v =
1
u
}=
1
maxu= (
mi=1
ci)1k
V d: 1) Gi s x, y, z l cc s dng tho mn x+y+z = 3. Chng minh
rng vi mi s k > 1 ta c :
xk + yk + zk 3
2) Tm gi tr b nht ca hm s:
u = x + y + z + t
trn min D = {x > 0, y > 0, z > 0, t > 0 : xy + xz +
xt+ yz + yt + zt = 1}.3) Cho n s dng xi ( i = 1,. . . ,n) c tng bng
a v k > 1. Chng minh
bt ng thc:
ni=1
xki ak
nk1
25
-
Gii. 1) p dng nh l 2.4.4 trn min D ta c:
xk + yk + zk
(x+ y + z)k 1
3k1 x
k + yk + zk
3k 1
3k1 xk + yk + zk 3
Du ng thc xy ra khi x = y = z = 1.
2) Ta c:
x2 + y2 + z2 + t2 = (x+ y + z + t)2 2(xy + xz + xt+ yz + yt+ zt)
= u2 2.
Theo nh l 2.4.4 (hoc theo bt ng thc Cauchy-Bunhiacovski) ta
c:
x2 + y2 + z2 + t2 14(x+ y + z + t)2 =
u2
4 u2 2 u
2
4 u2 8
3 u
8
3
Khi x = y = z = t =16ta c : xy+xz+xt+yz+yt+zt = 1 v u =
8
3. Vy:
minD
u =
8
3
3) p dng nh l 2.4.4 cho hm f =ni=1
xi trn min Rn+ ta c ngay:
ni=1
xki(ni=1
xi
)k 1nk1 ni=1
xki
(ni=1
xi
)knk1
=ak
nk1
Du ng thc xy ra khi tt c cc xi bng nhau v bng a/n.
By gi gi s (an) (n N) l mt dy cc s thc khng m nhng cha vhn s dng
sao cho chui dng
n=0
an hi t. Xt chui lu tha:
n=0
anxn (2.9)
Chui (2.9) hi t ti x = 1 nn theo nh l Abel, min hi t ca chui
(2.9) cha khong ng [0;1]. Gi tp cc im hi t dng ca chui (2.9)
l
26
-
X*, ta c 1X*. Gi f(x) l tng ca chui (2.9) trn min X*. p dng btng
thc (2.8) di dng chui ta c:
f(x) =
n=0
anxn (
n=0
an)1
1
k (
n=0
anxkn)
1
k f(x)
f(xk)
1
k
(n=0
an)1
1
k (2.10)
Trong ta xem v tri ca (2.10) bng 0 nu chui f(xk) =n=0
anxkn phn k (*). Bt ng thc (2.10) tng ng vi bt ng thc:
f(x)k
f(xk) (
n=0
an)k1 (x X) (2.11)
Khng kh khn c th chng minh rng nu ai0 l h s c ch s b nht
trong cc h s khc 0 ca chui (2.9) th: limx0+
f(x)k
f(xk)= ak1i0
Do nu xem ak1i0 l gi tr ca v tri bt ng thc (2.11) khi
x = 0 (**) th (2.11) ng c vi trng hp x = 0. Mt khc, rng:
f(1)k
f(1k)= f(1)k1 = (
n=0
an)k1
kt hp vi (2.11) ta suy ra:
max{f(x)k
f(xk): x X {0}} = f(1)k1 = (
n=0
an)k1
Nh vy ta thu c nh l:
nh l 2.4.5: Gi s chui dngn=0
an hi t v c tng dng. Gi f(x)
l tng ca chui lu than=0
anxn khi x thuc tp X cc im hi t khng
m ca chui . Khi ta c:
max{f(x)k
f(xk): x X} = f(1)k1 = (
n=0
an)k1
min{f(xk)
f(x)k: x X} = f(1)1k = (
n=0
an)1k
(ta gi nguyn cc quy c (*) v (**)).
27
-
V d : 1) Ta c : ex =n=0
xn
n!(x R) nn p dng nh l 2.4.5 ta c
ngay:
max{e2010x
ex2010 : x 0} = e2009 (k = 2010)
Nu nhn xt thm rnge2010x
ex2010
e2010x
ex2010 khi x 0 kt qu trn cn cho
khng nh mnh hn:
max{e2010x
ex2010 : x R} = e2009 (k = 2010)
2) Ta c: ln(1 x2) =
n=1
1
n(x
2)n (x (2; 2)) . p dng nh l 2.4.5 ta
c:
max{ln100(1 x
2)
ln(1 x100
2)
: 0 x < 1002} = (ln 2)99 ( k = 100).
2.5 Cc tr ca cc hm na cng tnh
Trong mc ny, ging nh u chng 1 i khi ta vit f(x) thay cho
f(x1, ..., xn), trong x = (x1, ..., xn) l mt vc t ca Rn. Ta cng
gi thit
rng hm f(x) xc nh trn tp con D ca Rn c tnh cht: x, y D ko theox
+ y D.
nh ngha 2.5.1: Hm f c gi l:
- Di cng tnh trn tp xc nh D ca n nu f c tnh cht:
f(x+y) f (x) + f(y)
vi mi cp vc t x, y D.- Trn cng tnh trn tp xc nh D ca n nu f c
tnh cht:
f(x+y) f (x) + f(y)
vi mi cp vc t x, y D.
28
-
Cc hm di cng tnh v trn cng tnh trn mt min no gi chung
l hm na cng tnh trn min .
V d: - Hm u = (ni=1
xki )1/k l hm di cng tnh trn min Rn+ vi k 1,
bi v theo bt ng thc Mincowski trn min Rn+ ta c:(ni=1
(xi + yi)k
)1/k(
ni=1
xki
)1/k+
(ni=1
yki
)1/k- Hm u = x2 l trn cng tnh trn khong (0, + ), bi v vi mi s
thc
x, y (0, + ) ta c:
(x+ y)2 = x2 + 2xy + y2 > x2 + y2
Mnh di y cho php ta thu c nhiu hm na cng tnh t mt
hm na cng tnh bit trc.
Mnh 2.5.1: 1) Gi s f(x) l hm mt bin xc nh trn khong
(0,+). Nu hm f(x)x
l hm khng tng th f(x) l hm di cng tnh trn
(0,+). Nu hm f(x)x
l hm khng gim th f(x) l hm trn cng tnh trn
(0,+).2) Gi s f(u) l hm di cng tnh v khng gim trn (0,+). Nu
u=g(x)
l hm nhn gi tr dng v di cng tnh trn Rn+ th hm hp f(g(x)) l
di cng tnh trn Rn+ .
3) Gi s f(u) l hm trn cng tnh v khng gim trn (0,+). Nu u=g(x)l
hm nhn gi tr dng v trn cng tnh trn Rn+ th hm hp f(g(x)) l
trn cng tnh trn Rn+ .
Chng minh. 1) Trc ht ta chng minh rng nu c, d l cc s dng th
vi cc s a, b tu ta c:
min(a
c,b
d) a+ b
c+ d max(a
c,b
d) (2.12)
Tht vy, khng gim tng qut gi sa
c bd. Khi c cc s m, n 0 sao
29
-
cho:a
c=bmd
,a+ n
c=b
d. Suy ra:
min(a
c,b
d) =
a
c=bmd
=a+ bmc+ d
a+ bc+ d
a+ n+ bc+ d
=a+ n
c=b
d= max(
a
c,b
d)
Vy (2.12) c chng minh.
By gi gi s x,y (0,+). V hm f(x)x
l khng tng trn (0,+) ta c:
f(x+ y)
x+ y min(f(x)
x,f(y)
y) f(x) + f(y)
x+ y f(x+ y) f(x) + f(y)
Vy hm f(x) l di cng tnh trn (0,+).Nu hm
f(x)
xl khng gim trn (0,+) v x,y (0,+) ta c:
f(x+ y)
x+ y max(f(x)
x,f(y)
y) f(x) + f(y)
x+ y f(x+ y) f(x) + f(y)
Vy hm f(x) l trn cng tnh trn (0,+).2) Gi s x, y l hai vc t tu ca
Rn+ . Vi cc gi thit nu trong khng
nh 2) ca mnh 2.5.1 ta c:
f(g(x+y)) f(g(x)+g(y)) f(g(x)) + f(g(y))
Vy hm f(g(x)) l di cng tnh trn Rn+ .
3) Gi s x, y l hai vc t tu ca Rn+ . Vi cc gi thit nu trong
khng
nh 3) ca mnh 2.5.1 ta c:
f(g(x+y)) f(g(x)+g(y)) f(g(x)) + f(g(y))
Vy hm f(g(x)) l trn cng tnh trn Rn+ .
V d: - Hm f(x) = ln(1 + x) l hm tho mnf(x)
xkhng tng trn
(0;+), cn bn thn f(x) thc s tng trn (0;+) do theo khng nh 2)ca
mnh 2.5.1 hm f(g(x)) l hm di cng tnh trn Rn+ vi mi hm g
di cng tnh trn Rn+ . Ni ring, hm u = ln(1 + (
ni=1
xki )1/k
)l di cng
tnh trn Rn+ .
30
-
- Trong Rn t: x1 =ni=1
|xi| vi mi vc t x = (x1, ..., xn) Rn. Vi micp vc t x, y Rn+ ta
c:
x+ y1 = x1 + y 1
Vy hm g(x) = x1 l cng tnh trn Rn+ (do g(x) va di cng tnh,va trn
cng tnh trn Rn+ ). Hm f(x) = x2 l trn cng tnh v thc s tng
trn (0,+ ) nn theo khng nh 3) ca mnh 2.5.1 ta c hm x21 l trncng
tnh trn Rn+ .
nh ngha 2.5.2: Hm n bin f(x) (x =(x1, ..., xn) D Rn) gi l
thunnht dng trn D nu D v f c cc tnh cht sau:
1) x =(x1, ..., xn) D ko theo tx D vi mi t > 0.2) f(tx) =
tf(x) vi mi t > 0 v mi x D.V d : - Hm u = (
ni=1
xki )1/k l thun nht dng trn min Rn+ vi mi
k>0. Khi k > 1 hm u l di cng tnh (bt ng thc Mincovski).
Khi 0 0}
=R2+ vi s thc n tu .
- Nu z = x l mt chun tu ca vc t x Rn+ th z va c tnh chtdi cng
tnh, va c tnh cht thun nht dng trn Rn, x1 l hm cngtnh v thun nht
dng trn Rn+ .
nh l 2.5.2: 1) Gi s w = f(u,v) l hm di cng tnh, thun nht
dng trn min R2+ v khng gim theo v vi mi u c nh. t:
G =
{x = (x1, ...xn) Rn+ :
ni=1
xi C = const > 0}
; z =ni=1
f(xi,1
xi)
Nu hm g(t) = f(t,1
t) khng tng theo t trn (0;C) th:
min {z : x = (x1, ..., xn) G}= f(C, n2
C)
31
-
2) Gi s w = f(u,v) l hm trn cng tnh, thun nht dng trn min
R2+
v khng tng theo v vi mi u c nh. t:
G =
{x = (x1, ...xn) Rn+ :
ni=1
xi C = const > 0}
; z =ni=1
f(xi,1
xi)
Nu hm g(t) = f(t,1
t) khng gim theo t trn (0;C) th :
max {z : x = (x1, ..., xn) G}= f(C, n2
C)
Chng minh. 1) Nuni=1
xi < C, thay x1 bi x1 > x1 > 0 sao cho x1 +
ni=2
xi=C. Do gi thit g(t) khng tng theo t (0;C) v hm f l di cng
tnhta c:
ni=1
f(xi,1
xi) f(x1,
1
x1) +
ni=2
f(xi,1
xi)
f(x1 +ni=2
xi,1
x1+
ni=2
1
xi) = f(C,
1
x1+
ni=2
1
xi)
Vy vi x = (x1, ..., xn) tu thuc G ta c:
z =ni=1
f(xi,1
xi) inf{f(C,
ni=1
1
xi) :
ni=1
xi = C, xi > 0}
Theo bt ng thc Cauchy Bunhiacovski, vi x = (x1, ..., xn) G
vni=1
xi = C ta c:
n2 (ni=1
xi)(
ni=1
1
xi) = C(
ni=1
1
xi) (
ni=1
1
xi) n
2
C
Bi v hm f khng gim theo bin th hai, ta suy ra khini=1
xi = C c bt
ng thc:
f(C,
ni=1
1
xi) f(C, n
2
C)
Vy: z f(C, n2
C) vi mi x = (x1, ..., xn) G.
Khi x1 = x2 = ... = xn =C
n, do tnh thun nht dng ca f ta c z=nf(
C
n,n
C)
= f(C,n2
C).
32
-
Vy min{z : x = (x1, ..., xn) G}= f(C, n2
C).
2) Nuni=1
xi < C, thay x1 bi x1 > x1 > 0 sao cho x1 +
ni=2
xi = C. Do gi
thit g(t) khng gim theo t (0;C) v hm f l trn cng tnh ta
c:ni=1
f(xi,1
xi) f(x1,
1
x1) +
ni=2
f(xi,1
xi)
f(x1 +ni=2
xi,1
x1+
ni=2
1
xi) = f(C,
1
x1+
ni=2
1
xi)
Vy vi x = (x1, ..., xn) tu thuc G ta c:
z =ni=1
f(xi,1
xi) sup{f(C,
ni=1
1
xi) :
ni=1
xi = C, xi > 0}
Theo bt ng thc Cauchy Bunhiacovski, vi x = (x1, ..., xn) G
vni=1
xi = C ta c:
n2 (ni=1
xi)(
ni=1
1
xi) = C(
ni=1
1
xi) (
ni=1
1
xi) n
2
C
Bi v hm f khng tng theo bin th hai, ta suy ra khini=1
xi = C c bt
ng thc:
f(C,
ni=1
1
xi) f(C, n
2
C)
Vy ta c: z f(C, n2
C) vi mi x = (x1, ..., xn) G.
Khi x1 = x2 = ... = xn =C
n, do tnh thun nht dng ca f ta c z=nf(
C
n,n
C)
= f(C,n2
C).
Vy max{z : x = (x1, ..., xn) G}= f(C, n2
C).
V d: 1) Trn R2+ hm f(u,v) = (uk + vk)1/k vi k > 1 l di cng
tnh
v thun nht dng, khng gim theo bin v (suy t bt ng thc
Mincovski
v biu thc ca f). Gi s:
G=
{x = (x1, ..., xn) Rn+ :
ni=1
xi 1}
33
-
Ta c:
g(t)= f(t,1
t
)= (tk +
1
tk)1/k
l hm thc s gim trn (0;1) bi v g(t) = (tk +1
tk)
1 kk (
t2k 1tk+1
) < 0 khi t (0;1). Nh vy, tt c cc gi thit v cc hm f, g trong
khng nh 1) ca nh
l 2.5.2 c tha mn. Theo khng nh 1) ca nh l ny ta c:
min
{ni=1
(xki +1
xki)1/k
:ni=1
xi 1, xi > 0 (i = 1, ..., n)}
= (1 + n2k)1/k
Khi k = 2, n =3 ta nhn c chng minh ca bi ton s V ca thi tuyn
sinh i hc khi A nm 2003:
Cho 3 s dng x,y,z tha mn x + y + z 1. Chng minh rng:x2 +
1
x2+
y2 +
1
y2+
z2 +
1
z282
2) Cho n s dng x1, ..., xn tha mnni=1
xi C. Tm gi tr ln nht ca biuthc:
z =ni=1
(xi 1xi)
Gii. Xt hm w = u v. Khi w l thun nht dng v cng tnh, do
c th coi w l trn cng tnh. Hm w thc s gim theo bin v vi mi u
c
nh. Hm g(t) = t 1tthc s tng trn (0;C) bi v g(t) = 1 +
1
t2> 0. Do
mi gi thit trong khng nh 2) ca nh l 2.5.2 c tho mn. Theo
khng
nh ny ta c:
max
{z =
ni=1
(xi 1
xi
):ni=1
xi C, xi > 0i}
= C n2
C
34
-
Chng 3
Cc tr ca cc hm ca hai athc i xng hai bin
3.1 Cc a thc i xng ca hai bin
nh ngha 3.1.1. Mt a thc i xng ca hai bin x, y l mt hm
dngmi=1
cixmiyni, trong ci l cc s thc, mi, ni l cc s nguyn khng m,
c tnh cht bt bin khi hon v cc k hiu x, y cho nhau.
V d: s1= x+ y, s2 = xy, n = xn + yn (n l s nguyn >1) l cc a
thc
i xng ca hai bin x, y.
Ta c khng nh sau (xem [Kurosh]):
Mnh 3.1.1: Mi a thc i xng ca hai bin x, y u l a thc ca
cc a thc i xng s1, s2.
Do mnh 3.1.1, mi hm ca hai a thc i xng ca hai bin x, y u
c th xem l hm ca hai bin s1 = x + y v s2 = xy. V vy ta s xt cc
hm
dng f(u,v), trong u = x + y, v = xy. Chng ny dnh cho bi ton tm
gi
tr ln nht v gi tr b nht ca cc hm dng f(x+y, xy) khi cc hm ny
c xc nh trn cc tp dng c bit ca R2. V cc kt qu lin quan xem
bi bo [H.V.Hng 2] .
35
-
3.2 Cc tr ca cc hm ca hai a thc i xng
ca hai bin
Trong mc ny k hiu ch mt trong cc khong (a;b), [a;b), (a;b],
[a;b], a 0 v b c th l + .nh l 3.2.1: Gi s f(u,v) l hm ca hai bin
u, v xc nh trn min
D =
{u < a; b >, v u
2
4
}v g(x,y) = f(x+y,xy) xc nh trn min H =
{(x, y) : u = x+ y < a, b >} R2. t N(x) =f (2x, x2), x
< a2;b
2>. Khi :
1) Nu trn min D =
{u < a, b >, v u
2
4
}hm f(u,v) khng tng theo bin
v vi mi u c nh v hm N(x) c gi tr b nht trn th:
min{g (x, y) : (x, y) H} = min{N (x) : x < a
2,b
2>
}.
2) Nu trn min D=
{u < a, b >, v u
2
4
}hm f(u,v) khng gim theo bin
v vi mi u c nh v hm N(x) c gi tr ln nht trn th:
max{g (x, y) : (x, y) H} = max{N (x) : x < a
2,b
2>
}.
Chng minh. 1) C nh mt gi tr u . Vi mi (x,y) H thamn x + y = u ta
c:
xy (x+ y)2
4=u2
4(3.1)
V hm f(u,v) khng tng theo v, vi (x,y) H tha mn x + y = u ta
c:
g(x, y) = f(x + y, xy) = f(u, xy) f(u,u2
4) (3.2)
Khi x = y = u/ 2 ta c (x,y) H v:
g(x,x)=f(2x,x2)=N(x)=f(u,u2
4) (3.3)
V hm N(x) c gi tr b nht trn , tn ti im x*< a
2;b
2> sao
cho:
36
-
min
{N (x) : x < a
2;b
2>
}=N(x*) (3.4)
T (3.2), (3.3), (3.4) ta suy ra:
g(x, y) N(x*) = f(2x, x
2)vi mi (x, y) H.
Ly u* = 2.x* v x = y = x* ta c (x, x) H v g(x*, x*) = N(x*).
Vy:
min{g (x, y) : (x, y) H}=min{N (x) : x < a
2;b
2>
}= N(x*)
2) Nu f(u,v) l hm khng gim theo v th t (3.1) ta suy ra vi mi
(x,y)
tho mn x+y = u ta c:
g(x, y) = f(x + y, xy) = f(u, xy) f(u,u2
4) (3.5)
Khi x = y = u/ 2 ta c (x,y) H v:
g(x,x)=f(2x,x2)=N(x)=f(u,u2
4) (3.6)
V hm N(x) c gi tr b nht trn , tn ti im x** < a
2;b
2>
sao cho:
max
{N (x) : x < a
2;b
2>
}= N (x) (3.7)
T (3.5), (3.6), (3.7) ta suy ra:
g(x, y) N (x) =f(2x, x
2)vi mi (x,y) H.
Ly u** = 2.x** v x = y = x** ta c (x**,x**) H v g(x**, x**) =
N(x**).Vy:
max{g (x, y) : (x, y) H} = max{N (x) : x < a
2;b
2>
}= N(x**).
nh l 3.2.1 c th p dng gii mt lot cc bi ton v tm gi tr b
nht, gi tr ln nht ca cc hm ca hai a thc i xng ca hai bin x,
y
trn min dng H. Nhng bi ton ny thng xut hin trong cc k thi
tuyn
37
-
sinh i hc v cc k thi hc sinh gii ton bc trung hc c s hoc trung
hc
ph thng. Di y l mt s v d:
V d 1: Tm gi tr nh nht ca biu thc A = x3+y3+3(xy1)(x+y2),trong
x, y l cc s thc tho mn (x 4)2 + (y 4)2 + 2xy 32.
( thi tuyn sinh i hc nm 2012, mn Ton-khi D).
Gii. Ta c: (x 4)2 + (y 4)2 + 2xy 32 (x+ y)2 8(x + y) 0 0 x+ y
8;
A = (x+ y)3 3(x+ y) 6xy + 6
Vy bi ton c dng: Tm gi tr b nht ca hm f(u,v) = u3 3u 6v+6,trong
u = x + y, v = xy trn min H ={(x, y) : 0 x+ y 8}.
Hm f(u,v) r rng gim theo bin v = xy v min H c dng nh trong
nh
l 3.2.1 (khong trong trng hp ny l khong ng [0;8]). p dng
khng nh 1) ca nh l ta c:
min{A : 0 x+ y 8} = min{N (x) = 8x3 6x2 6x+ 6 : 0 x 4}.Ta c :
N(x) = 24x2 12x 6, N(x) = 0 x = 1
5
4. V xt trn min
0x4 nn ta ch ly gi tr x = 1 +5
4. So snh cc gi tr N(0) = 6, N(
1 +5
4)
=17 55
4, N(4) = 398 ta suy ra:
min{A : 0 x+ y 8}=min{N (x) = 8x3 6x2 6x+ 6 : 0 x 4}=17 554
.
V d 2: Cho cc s thc x, y thay i v tho mn (x+ y)3 + 4xy 2.Tm gi
tr nh nht ca biu thc A = 3
(x4 + y4 + x2y2
) 2 (x2 + y2)+ 1.( thi tuyn sinh i hc nm 2009, mn Ton - khi
B).
Gii. t H ={(x, y) : (x+ y)3 + 4xy 2}. Bi v (x+ y)2 4xy 0 nn
ta
suy ra:
(x+ y)3+ (x+ y)2 - 2 0 x + y 1
Nh vy min H nm trong min G = {(x, y) : (x+ y) 1}. Do :
38
-
inf{A : (x, y) H} inf{A : (x, y) G} (3.8)
Ta c : A = 3(x+ y)4 12xy(x+ y)2 + 9x2y2 - 2(x+ y)2 + 4xy + 1. t
u =
x + y, v = xy ta nhn c:
A = f(u,v) = 3u4 12vu2 + 9v2 2u2 + 4v + 1
f
v= 12u2 + 18v + 4 = 4 (1 u2)+ 18(v 4
9u2)
Trn min D=
{u 1, v u
2
4
}ta c
f
v< 0, do f(u,v) gim theo v vi
mi u 1 c nh. t N(x) = f(2x, x2) = 9x4 4x2 + 1 ta c:N(x) = 36x3
8x > 0 trn min x 1
2
Do : min{N (x) : x 1
2
}= N
(1
2
)=
9
16. Theo khng nh 1) ca nh l
3.2.1 ta c:
min{A : (x, y) G}=min{N (x) : x 1
2
}=
9
16
Do (3.8) ta c: inf {A : (x, y) H} 916
. Nhng vi x = y = 1/2 ta c
(x+ y)3 + 4xy = 2 vy im(1
2;1
2
) H v A
(1
2;1
2
)=
9
16nn ta suy ra
min{A : (x, y) H} = 916
V d 3: Tm gi tr ln nht ca hm s:
z =x3 + y3
x3 + y3 xytrn min H = {(x, y) : x+ y a > 1}.
Gii. Nu xy < 0 ta c: x3 + y3 xy = (x+ y) (x2 + y2 xy) xy a(x2
+ y2 xy) xy > 0. Nu xy 0 ta c x3 + y3 = (x + y)(x2 + y2 xy)
a(x2 + y2 xy) > axy xy trn min H. Vy hm z c ngha trn H. t u =
x+ y , v = xy ta c:
z =(x+ y)3 3xy(x+ y)
(x+ y)3 3xy(x+ y) xy=
u3 3uvu3 3uv v = f(u,v)
39
-
f
v=
u3
(u3 3uv v)2> 0 khi u a > 1.
Vy hm f(u,v) khng gim theo v trn min G =
{(u, v) : v u
2
4, u a > 1
}vi mi u c nh. Theo khng nh 2) ca nh l 3.2.1 ta c:
max{z : (x, y) H}= max{N (x) = f
(2x, x2
): x a
2
}.
Nhng N(x) =2x3
2x3 x2 =2x
2x 1 = 1 +1
2x 1 1 +1
a 1 vi mi xa
2. Khi
x =a
2ta c N(
a
2) = 1 +
1
a 1 .
Vy max{z : (x, y) H}= max{N (x) = f
(2x, x2
): x a
2
}= 1 +
1
a 1 .nh l 3.2.2: Gi s f(u,v) l hm ca hai bin u, v xc nh trn
min
D=
{(u, v) : u 0, v < a; b >, v u
2
4
}v g(x,y) = f(x+y, xy) xc nh trn min
H = {(x, y) : v = xy < a; b >} R2. t N(x) = f(2x, x2), x
< a;b > . Khi:
1) Nu trn min D =
{(u, v) : u 0, v < a, b >, v u
2
4
}hm f(u,v) khng
gim theo bin u vi mi v c nh v hm N(x) c gi tr b nht trn min{x
< a;b >} th:
min{g (x, y) : (x, y) H}= min {N (x) : x < a;b >}.2) Nu
trn min D =
{(u, v) : u 0, v < a, b >, v u
2
4
}hm f(u,v) khng
tng theo bin u vi mi v c nh v hm N(x) c gi tr ln nht trn min{x
< a;b >} th:
max {g (x, y) : (x, y) H}= max {N (x) : x < a;b >}.Chng
minh. 1) C nh mt gi tr v < a;b >. Vi mi (x,y) H tha
mn xy = v ta c:
v = xy (x+ y)2
4 x+ y 2v (3.9)
40
-
V hm f(u,v) khng gim theo u, vi (x,y) H tha mn xy = v ta c:
g(x, y) = f(x + y, xy) = f(x + y, v) f(2v, v) (3.10)
Khi x = y =v ta c (x,y) H v:
g(x,x)=f(2x, x2
)=N(x)=f(2
v, v) (3.11)
Gi s:
min{N (x) : x < a;b >} = N(x*); (x*< a;b > )
(3.12)
T (3.10), (3.11), (3.12) ta suy ra:
g(x,y) N(x*) = f(2x, x
2)vi mi (x,y) H.
Ly v* =x2
v x = y = x* ta c (x*,x*) H v g(x*, x*) = N(x*). Vy:
min{g (x, y) : (x, y) H}= min {N (x) : x < a;b >}=
N(x*).2) Nu f(u,v) l hm khng tng theo u th t (3.9) ta suy ra vi mi
(x,y)
tho mn xy = v > 0 ta c:
g(x, y) = f(x + y, xy) = f(x + y, v) f(2v, v) (3.13)
Khi x = y =v ta c (x,y)H v:
g(x,x)=f(2x, x2
)=N(x)=f(2
v, v) (3.14)
Gi s :
max{N (x) : x < a;b >} = N(x**); (x**< a;b > )
(3.15)
T (3.13), (3.14), (3.15) ta suy ra:
g(x,y) N(x**) = f(2x, x
2)vi mi (x,y)H
Ly v** = x2
v x = y = x** ta c (x**,x**)H v g(x**, x**) = N(x**).Vy:
41
-
max {g (x, y) : (x, y) H}= max {N (x) : x < a;b >} =
N(x**)nh l 3.2.2 c chng minh hon ton.
Ta a ra mt s v d minh ho cho nh l 3.2.2.
V d 4: Tm gi tr b nht trn min H ={(x, y) : xy 1
9, x > 0, y > 0
}ca hm s:
z =x3 + y3 xyx3 + y3 + 2xy
Gii. t u = x + y, v = xy, ta c :
z =(x+ y)3 3(x+ y)xy xy(x+ y)3 3(x+ y)xy + 2xy
=u3 3uv vu3 3uv + 2v
z
u=
9(u2 v)v(u3 3uv + 2v)2
Trn min D =
{(u, v) :
1
9 v u
2
4
}ta c
z
u> 0 do hm z tng theo
bin u. Theo khng nh 1 ca nh l 3.2.2 ta c:
min{z : (x, y) H} = min{N (x) =
2x3 x22x3 + 2x2
=2x 12(x+ 1)
: x 13
}nu v phi tn ti. Hm N(x) c o hm N(x) > 0 trn min x 1
3, nn ta
c: min{N (x) : x 1
3
}= N(
1
3) = 1
8. Vy:
min
{Z =
x3 + y3 xyx3 + y3 + 2xy
: (x, y) H}
= 18.
V d 5: Tm gi tr ln nht ca hm s:
z=x4 + y4 + xy(x2 + y2)
2(x4 + y4) xy(x2 + y2)trn min H = {(x, y) : x > 0, y >
0}.Gii. Ta c : 2
(x4 + y4
)xy (x2 + y2)= (x2+y2)[(x y2)2+
3y2
4]+(x2 y2)2 > 0
khi x > 0, y > 0 do z hon ton c xc nh trn min H. t u = x +
y,
v = xy ta c biu din:
42
-
z =u4 3vu2
2u4 9vu2 + 6v2
Ly o hm z theo bin u ta c:
z
u= 6uv
(u4 4vu2 + 6v2)
(2u4 9vu2 + 6v2)2
Trn min D =
{(u, v) : u > 0, 0 < v u
2
4
}ta c
z
u< 0, do z l hm gim
theo u vi mi v c nh. Theo khng nh 2 ca nh l 3.2.2 ta c:
max
{z =
x4 + y4 + xy(x2 + y2)
2(x4 + y4) xy(x2 + y2) : x > 0, y > 0}
= max
{N (x) =
4x4
2x4= 2 : x > 0
}= 2.
43
-
Kt lun
Bn lun vn Cc tr ca mt s hm nhiu bin c cc dng c bit
a ra cng thc tnh gi tr ln nht hoc gi tr b nht ca mt s hm
nhiu
bin c dng c bit. Cc hm nhiu bin c dng c bit ny bao gm: cc
phn thc k- chnh quy, cc hm c dng t s ca hai phn thc ng dng,
cc hm biu din qua cc hm na cng tnh, cc hm ca hai a thc i xng
ca hai bin. Mt s cc kt qu c tc gi tng kt t cc ti liu tham
kho,
cc kt qu khc do tc gi t chng minh di s hng dn ca TS. Hong
vn Hng, Vin Khoa hc C bn, i hc Hng hi Vit nam (cc khng nh
v cc phn thc k chnh quy khi k 6= 0 ca nh l 2.1.1, nh l 2.5.2 v
ccnh l 3.2.1, 3.2.2). Cc v d minh ha cho cc nh l trong bn lun vn
ch
yu c ly t cc thi tuyn sinh i hc mn Ton ca cc khi A, B, D
t nm 2002 n nm 2012 v mt s sch tham kho v bt ng thc. iu
ny chng t cc kt qu ca bn lun vn l hu ch khi gii mt s cc bi
ton v gi tr b nht v ln nht trong chng trnh ph thng.
Cc kt qu ca chng 3 cn c th m rng cho cc hm ca cc a thc
i xng vi s bin ln hn. tin li cho vic din t, trong bn lun vn
tc gi s dng cc thut ng v k hiu thuc phm vi ca chng trnh
ton cao cp. Tuy nhin, cc l lun trong cc chng minh hon ton s cp.
V
vy bn lun vn ny hon ton thuc chuyn ngnh Phng php Ton s cp
v c th c s dng lm ti liu ging dy cho gio vin cng nh lm ti
liu hc tp cho hc sinh t bc trung hc c s tr ln trong cc chng
trnh
nng cao v bt ng thc v cc bi ton cc tr.
44
-
Ti liu tham kho
[1]. [H.V.Hng 1] Hong Vn Hng (2010), "Cc tr ca mt lp cc hm c
dng t s ca hai hm i s", Tp ch Khoa hc Cng ngh Hng hi (ISSN
1859 -316X), s 24-11/2010, trang 92-97.
[2]. [H.V.Hng 2] Hong Vn Hng (2013), "Phng php hm tri v hm
non tm gi tr ln nht v b nht ca mt s hm ca hai a thc i xng",
Tp ch Khoa hc Cng ngh Hng hi (ISSN 1859 -316X), s 33-01/2013,
trang
97-101.
[3]. [P.H.Khi 1] Phan Huy Khi (1998), "Cc tiu ca phn thc chnh
quy",
Tuyn tp 30 nm Tp ch Ton hc v Tui tr, Nh xut bn Gio dc,
(trang
231- 234).
[4]. [P.H.Khi 2] Phan Huy Khi (1998), 10.000 bi ton s cp - Bt
ng
thc hnh hc, Nh xut bn H Ni.
[5]. [Tr.Phng] Trn Phng (1997), Bt ng thc, Tp 1. Nh xut bn
thnh ph H Ch Minh.
[6]. [Hardy, Littllewood,Polya] .G.H.Hardy, J.E.
Littllewood,G.Polya (1981),
Bt ng thc, Nh xut bn i hc v trung hc chuyn nghip (dch t bn
ting Nga).
[7]. [Kurosh] A.G Kurosh (1975), Gio trnh i s cao cp, Nh xut
bn
Khoa hc, Moskva (ting Nga).
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