1. ## Nice Inequality !!

Hii !

a, b and c the réels posifives numbers , prove That :

$\displaystyle a^2+b^2+c^2 \geq a+b+c+2( \frac{a-1}{a+1} + \frac{b-1}{b+1} + \frac{c-1}{c+1} )$

Have Fun

2. Originally Posted by Perelman
Hii !

a, b and c the réels posifives numbers , prove That :

$\displaystyle a^2+b^2+c^2 \geq a+b+c+2( \frac{a-1}{a+1} + \frac{b-1}{b+1} + \frac{c-1}{c+1} )$

Have Fun
Spoiler:
Take all the terms over to the left side:

$\displaystyle a^2 - a -2\Bigl(\frac{a-1}{a+1}\Bigr) = \frac{(a-1)^2(a+2)}{a+1}$, which is obviously positive if a is! – and similarly for the b and c terms.

3. Hii!!

Perfect Mr Opalg !

So My Solution :
$\displaystyle a^2+b^2+c^2 \geq a+b+c+2( \frac{a-1}{a+1} + \frac{b-1}{b+1} + \frac{c-1}{c+1} )$
$\displaystyle \Leftrightarrow$ $\displaystyle a^2+ b^2+ c^2 + 4(\frac{1}{a+1} + \frac{1}{b+1}+\frac{1}{c+1}) \geq a +b +c + 6$

Put : $\displaystyle a+1=x$ , $\displaystyle b+1=y$ and $\displaystyle c+1=z$ :
$\displaystyle x^2+y^2+z^2+4(\frac{1}{x} + \frac{1}{y}+\frac{1}{z}) \geq 3(x+y+z)$
By Cauchy Schwartz :
$\displaystyle \frac{1}{x} + \frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z}$

And We Have : $\displaystyle x^2+ y^2+ z^2 \geq \frac{(x+y+z)2}{3}$

So, We need prove : $\displaystyle \frac{(x+y+z)^2}{3} + \frac{36}{x+y+z} \geq 3(x+y+z)$

$\displaystyle \Leftrightarrow$ $\displaystyle (x+y+z)^2 -9(x+y+z)^2+108 \geq 0$
$\displaystyle \Leftrightarrow$ $\displaystyle (x+y+z-6)^2 (x+y+z+3) \geq 0$

Cqfd .

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