algebra inequality

**prasum** · November 18th 2010, 09:53 AM

if a>0,b>0,c>0 then prove that a/(b+c)+b/(a+c)+c/(a+b)>=3/2

**red_dog** · November 20th 2010, 11:47 PM

Let $b+c=x, \ a+c=y, \ a+b=z$

Then $a=\dfrac{-x+y+z}{2}, \ b=\dfrac{x-y+z}{2}, \ c=\dfrac{x+y-z}{2}$

The inequality becomes

$\dfrac{-x+y+z}{2x}+\dfrac{x-y+z}{2y}+\dfrac{x+y-z}{2z}\geq \dfrac{3}{2}$

or $\dfrac{1}{2}\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfra c{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\df rac{z}{y}\right)-3\right]\geq\dfrac{3}{2}$

But it is well known that if $a>0$ then $a+\dfrac{1}{a}\geq 2$

Can you finish?

**Krizalid** · November 21st 2010, 04:25 AM

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