Inequality

great_math · #1 $Old$ 10-29-2008, 10:04 AM

Prove that for all $x,y,z>0$

$\frac{x}{(x+y)(x+z)}+\frac{y}{(y+z)(y+x)}+\frac{z}{(z+x)(z+y)}\le \frac{9}{4(x+y+z)}$

NonCommAlg · #2 $Old$ 10-30-2008, 02:59 AM

Quote:

Originally Posted by great_math $View Post$

Prove that for all $x,y,z>0$

$\frac{x}{(x+y)(x+z)}+\frac{y}{(y+z)(y+x)}+\frac{z}{(z+x)(z+y)}\le \frac{9}{4(x+y+z)}$

so we want to prove that: $8(x+y+z)(xy+yz+zx) \leq 9(x+y)(y+z)(z+x),$ which after simplifying becomes: $6xyz \leq x^2y+yz^2+x^2z +y^2z + xy^2+ xz^2 . \ \ \ \ (1)$

dividing both sides of (1) by $xyz$ gives us: $6 \leq \frac{x}{z} + \frac{z}{x} + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y},$ which is clearly true by AM-GM, or (if you don't like AM-GM) because: $\forall t > 0 : \ t + \frac{1}{t} \geq 2. \ \ \Box$