Three-variable inequality

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Three-variable inequality

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Source: 2017 Belarus Team Selection Test 4.2

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#1 h Mar 31, 2019, 6:09 PM • 2 Y

Y by NoDealsHere, Adventure10

Given that $x$ , $y$ , $z$ are positive real numbers satiafying $x+y+z=2$ , prove the inequality
$\frac{(x-1)^2}{y}+\frac{(y-1)^2}{z}+\frac{(z-1)^2}{x}\geqslant\frac14\left(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x} \right).$

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arqady

#2 h Apr 1, 2019, 3:28 AM • 4 Y

Y by NoDealsHere, ehuseyinyigit, Adventure10, Mango247

Because $\frac{(x-1)^2}{y}+\frac{(y-1)^2}{z}\geq\frac{(x+y-2)^2}{y+z}=\frac{z^2}{y+z}.$

This post has been edited 1 time. Last edited by arqady, Nov 9, 2020, 5:23 AM

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#3 h Aug 20, 2022, 5:43 PM

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:welcome:

By T2 lemma we can write :

$\sum\frac{(x-1)^2}{y} \geq \frac{1}{2}\sum \frac{(x+y-2)^2}{y+z} = \frac{1}{2}\sum \frac{z^2}{y+z}$ Now we need prove : $2\sum \frac{z^2}{y+z} \geq \frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}$
Then .......?! :-D

:-D

:wow:

$\blacksquare$

This post has been edited 1 time. Last edited by strong_boy, Aug 20, 2022, 5:44 PM

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#4 h Aug 21, 2022, 10:46 AM • 2 Y

Y by strong_boy, ehuseyinyigit

@above well $2z^2 - (y^2 + z^2) = z^2 - y^2$ , so you want $\sum \frac{z^2-y^2}{y+z} = \sum (z-y) = 0$ to be at least $0$ , done.

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#5 h Sep 17, 2023, 5:12 PM

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This can be generalized for $x,y,z\in \mathbf{R^+}$ such that $x+y+z=2k$

$\sum_{cyc}{\frac{(x-k)^2}{y}}\geq \frac{1}{4}\left(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}\right)$

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#6 h Sep 17, 2023, 5:12 PM

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Arqady's solution below destroy it in a minute.

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