Cool Inequality 10 Trouble some

Algebra Level 4

${\sqrt { \left( xy+yz+zx \right) \left( \frac { 1 }{ \left( x+y \right) ^{ 2 } } +\frac { 1 }{ \left( y+z \right) ^{ 2 } } +\frac { 1 }{ { \left( z+x \right) }^{ 2 } } \right) } }$

If $x,y$ and $z$ are positive reals, find the minimum value of the expression above.

The answer is 1.50000.

4 solutions

Rishabh Jain
Jan 26, 2016

Is this correct??

This does not actually show that 1.5 is the minimum.

For example, say we have a non-negative real number $x$ and we are looking for the minimum of $x+3$ . Using the same logic as your solution, we have: $x+3\leq x+5$

$x+5\geq5$

So the minimum of $x+3$ is $5$ .

This is clearly untrue.

Mark Gilbert - 5 years, 4 months ago

Equality condition is important, in my solution equality at any step implies equality at all steps. What you stated x+3 $\leq$ x+5 has equality when x+3=x+5 or 3=5 which is absurd!!

Rishabh Jain - 5 years, 4 months ago

Okay, new example.

We have non-negative real numbers $x$ and $y$ and we are looking for the minimum of $\frac{4xy+1}{x^2+y^2+1}$ .

By AM-GM, $\frac{2x^2+2y^2+1}{x^2+y^2+1}\geq\frac{4xy+1}{x^2+y^2+1}$ with equality when $x=y$ .

$\frac{2x^2+2y^2+1}{x^2+y^2+1}\geq1$

So the minimum of $\frac{4xy+1}{x^2+y^2+1}$ is $1$ .

This is untrue because the minimum of $\frac{4xy+1}{x^2+y^2+1}$ does not occur when $x=y$ .

For your proof to actually work, you must now prove that the minimum of the original problem occurs when $x=y=z$ .

Mark Gilbert - 5 years, 4 months ago

Son Nguyen
Jan 30, 2016

This problem was in Iran Mathematics Olympiad 1996 and my solution is applying Schur inequality. Prove that A= $\sqrt{(xy+yz+xz)[\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}]}\geq \frac{3}{2}$ Or $A^{2}\geq \frac{9}{4}$

Rewrite the polynomial we have: $\sum 4x^5y-x^4y^2-3x^3y^3+x^4yz-2x^3y^2z+x^2y^2z^2\geq 0$ By applying Schur inequality we have: $x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y)\geq 0$ $\Leftrightarrow \sum x^4yz-2x^3y^2z+x^2y^2z^2\geq 0-(1)$ By applying AM-GM we have: $\sum (x^5y-x^4y^2)+3(x^5y-x^3y^3)\geq 0-(2)$ From (1)+(2) we have the solution

Prince Loomba
Jan 26, 2016

I also put x=y=z=1

Mark Vincent Esmeralda Mamigo
Jan 26, 2016

For minimum value, let x=y=z=1

Comeon lakshay did you seriously had to write upto 5 decimal places?

Shreyash Rai - 5 years, 4 months ago

I think... to confuse people!

Prince Loomba - 5 years, 4 months ago

Of course and please tag me like this @Shreyash Rai so we can talk.

Department 8 - 5 years, 4 months ago

neat trick K thats what ill do @Lakshya Sinha .

Shreyash Rai - 5 years, 4 months ago

See ,told u let it gain level.Simple cauchy one in the end simplifies to 1/2+1/2+1/2

Right?

Kaustubh Miglani - 5 years, 4 months ago

@Kaustubh Miglani – Post the solution

Department 8 - 5 years, 4 months ago

@Department 8 – IDK latex.In fiitjee i will explain method and u can lateify for me,ok

Kaustubh Miglani - 5 years, 4 months ago

Why must x=y=z=1?

Pi Han Goh - 5 years, 4 months ago

That's my intuition on minimum values.. :) sorry i got no proof, hehehe..

Mark Vincent Esmeralda Mamigo - 5 years, 4 months ago

I solved it by applying AM-GM 2 times... And condition of inequality came out to be x=y=z $\neq$ 0 .Previously I had also posted my solution only to delete it accidently and now I am not allowed to post a solution. :(

Rishabh Jain - 5 years, 4 months ago

you can also solve it by using Cauchy Schwartz inequality. But since im feeling lazy ill let someone else post the solution.

Shreyash Rai - 5 years, 4 months ago

@Shreyash Rai – I'll latexify my solution(if it's correct) as soon as I'll get time

Rishabh Jain - 5 years, 4 months ago

Cool Inequality 10 Trouble some

The answer is 1.50000.

4 solutions

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