9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum ­ Calvin Lin | Brilliant

https://brilliant.org/problems/find­tuongs­minimum/?group=3WveLY4nD4xV&ref_id=967118 1/5

Let , and be positive real numbers such that . What is the minimum value of

This problem is posed by Tuong N.

Need help? See related wiki

Share it. 39% of people got it right

Correct!You answered 64.

11 Solutions Vote up solutions you admire

From the HM-AM (harmonic mean-arithmetic mean) inequality, we can see that

But the LHS of the inequality is just , so we have . The

inequality holds if and only if , i.e. when . So the minimumof is indeed attainable, and our answer is .

Reply Subscribe to comments

Apply the Cauchy-Schwarz's inequality . If ;

and , then the equality holds. So the mininum value of is 64.

Reply

Find Tuong's minimum Algebra  Level 4    165 points

Hide 11 solutions

Derek Khu 1 year, 4 months ago

14

Hưng Hồ 1 year, 4 months ago

11

Home

Stats

Profile

Interact

Community

Publish

About

Topics

Algebra

Geometry

Number Theory

Calculus

Combinatorics

Basic Mathematics

Logic

Classical Mechanics

Electricity and Magnetism

Computer Science

Quantitative Finance

SAT® Math

JEE Math

by Calvin Lin609 Solvers

Solution writing guide: Level 4Formatting guide

Preview

Krutarth Patel 

Write a solution.

Insert an image

Like Repost Save Share More

Search

9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum ­ Calvin Lin | Brilliant

https://brilliant.org/problems/find­tuongs­minimum/?group=3WveLY4nD4xV&ref_id=967118 2/5

From the AM-GM inequality, for any positive reals , note

with equality attained when . Now since ,

with equality attained if

With the added constraint that , we find

satisfies the conditions and achieves the minimum, which is .

Reply

By Cauchy Inequality we have (1/x+1/y+4/z+16/t)*(x+y+z+t)>=(sqrt(1)+sqrt(1)+sqrt(4)+sqrt(16))^2=(1+1+2+4)^2=8^2=64

Reply

Let the required expression be .

If we are able to find numbers, such that their sum is , i.e. 1; and the sum of theirreciprocals is , i.e. ; then using the inequality we can find the

minimum value of P. Here is the arithmetic mean of the numbers, and is the harmonicmean of the numbers.

This means that for each and , we have to find values of and such that is thecoefficient of and , and is the coefficient for . is the no. of times term is repeated.

The following equation is obtained for .

Hero P. 1 year, 4 months ago

10

Mihai Nipomici 1 year, 4 months ago

5

Pranshu Gaba 1 year, 4 months ago

5

9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum ­ Calvin Lin | Brilliant

https://brilliant.org/problems/find­tuongs­minimum/?group=3WveLY4nD4xV&ref_id=967118 3/5

and

and

Similarly, we get the following equations for and

and

and

and

So, we get the following terms,

.

The of these numbers is

The of these numbers is

Using the inequality, we get

Thus, we can conclude that the minimum value of , i.e. is 64.

Reply

{(1/√x)^2+(1/√y)^2+(2/√z)^2+(4/√t)^2} {(√x)^2+(√y)^2+(√z)^2+(√t)^2}≥{(1/√x)(√x)+(1/√y)(√y)+(2/√z)(√z)+(4/√t)(√t)}^2 (1/x+1/y+4/z+16/t)(1)≥(1+1+2+4)^2(1/x+1/y+4/z+16/t)≥(8)^2 (1/x+1/y+4/z+16/t)(1)≥64. Done.Sorry i do not know how to use latexXD.

Reply

By Cauchy-Schwarz, we have (x+y+z+t)(1/x+1/y+4/z+16/t)>=(1+1+2+4)^2=64. We know thatx+y+z+t=1, so we have 1/x+1/y+4/z+16/t>=64.

Reply

Joshua Richard Theodoroes 1 year, 4 months ago

5

Kevin Chang 1 year, 4 months ago

5

Patrick Corn 1 year, 4 months ago

2

9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum ­ Calvin Lin | Brilliant

https://brilliant.org/problems/find­tuongs­minimum/?group=3WveLY4nD4xV&ref_id=967118 4/5

The minimum must be a local minimum on the hyperplane in 4-space. It's not onthe boundary of the region in question, because the quantity actually increases withoutbound as it approaches that boundary.

To find the local minima, we use Lagrange multipliers; they say that should be parallel to . Since all the variables are positive,

we get . This quickly leads to Since there is onlyone extremum in the region, it's either a global minimum or a global maximum, and it's clearly aglobal minimum because the expression blows up at the boundary. Plugging in, we get an answer of

.

Reply

Considere a funcao f(x,y,z,t)=1/x + 1/y + 4/z + 16/t Existe uma restricao: x+y+z+t=1 Derivef(x,y,z,t) com relação a apenas uma das variaveis e iguale a uma constante L, multiplicada a derivadade g(x,y,z,t) com relação apenas a essa mesma variável Para x, temos: A derivada de f apenas emrelacao a x é a derivada de 1/x: -1/x² A derivada de g apenas em relação a x é a derivada de x: 1 Logovoce escreve: -1/x² = L1 Faca o mesmo para todas as variaveis, utilizando o mesmo L: -1/y² = L1-4/z² = L1 -16/t² = L1 Chame L de -1/c^2 -1/x² = L = -1/c² => x = c -1/y² = L = -1/c² => y= c -4/z²= L = -1/c² => z = 2c -16/t² = L = -1/c² => t = 4c Agora, use o fato de x+y+z+t = 1 pra achar c.c+c+2c+4c = 1 8c=1 c=1/8 Assim, temos: x=1/8 y=1/8 z=21/8=1/4 t=41/8=1/2 f(1/8,1/8,1/4,1/2) =8+8+44+162=64

Reply

x<=y<z<t and we get x=1/8 y=1/8 z=1/4 t=1/2

1/x + 1/y + 4/z + 16/t = 8 + 8 + 16 + 32 = 64

Reply

Once we have and , and x,y,z,t belong to , we may conclude

that:

I- If one of them belonged to , the sum would not be 1. So, they belong to .

II - Since they're fractions, they may be written as .

III- . And .

IV- To reach the lowest sum, must be the lowest possible. For this, must be the lowest. So,

.Then, .

V - So .

VI- .

Pedro Mourão 1 year, 4 months ago

2

Syifa Adelia 1 year, 4 months ago

0

Marciel Silva de Almeida 1 year, 4 months ago

0

9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum ­ Calvin Lin | Brilliant

https://brilliant.org/problems/find­tuongs­minimum/?group=3WveLY4nD4xV&ref_id=967118 5/5

Using the same metod, we may infer, firstly, that would be . But, we have to consider that for , must be the lowest possible, and, in this case( ), .

Obviously, we may say that , because different numbers would make that would

make for every variable( ).

When we affirm that , . Then .

So, the minimum number is 64.

Reply

Try related problems

Minimum valueAlgebra  Level 4 

Shhhh!!!! Don't give me the hint....Algebra  Level 4 

Find the MinimumAlgebra  Level 4 

Minima doesn't always involve symmetry!Algebra  Level 4 

Terms   Privacy  © Brilliant 2015

Practice math   Practice math