INEQUALITY'S BLOG

July 29, 2010

(van khea):Problem 022

Filed under: Uncategorized — KKKVVV @ 1:11 am

Let \displaystyle x, y, z>0; 2(xy+yz+zx)\geq \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}. Prove that:

\displaystyle \frac{(x+y)^{3}}{1+2xy}+\frac{(y+z)^{3}}{1+2yz}+\frac{(z+x)^{3}}{1+2zx}\geq 8

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