# Nice Inequality !!

• February 4th 2010, 09:12 AM
Perelman
Nice Inequality !!
Hii !

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

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

Have Fun (Wink)
• February 5th 2010, 01:04 PM
Opalg
Quote:

Originally Posted by Perelman
Hii !

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

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

Have Fun (Wink)

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

$a^2 - a -2\Bigl(\frac{a-1}{a+1}\Bigr) = \frac{(a-1)^2(a+2)}{a+1}$, which is obviously positive if a is! – and similarly for the b and c terms.
• February 6th 2010, 02:19 AM
Perelman
Hii!!

Perfect Mr Opalg !

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

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

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

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

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

Cqfd .
• February 6th 2010, 03:08 AM
mr fantastic
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