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Numbers $x,y,z$ satisfy $x\in(0,1], y\in(0,1], z\in(0,1]$. Prove inequality: $$\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le \frac{x+y+z}{x+y+z+xyz}$$

My work so far:

$\frac x{2+xy+yz}\le \frac x{x^2+xz+xy+yz}=\frac x{(x+z)(x+y)}$

Then $$\frac x{2+xy+yz}+\frac y{2+yz+zx}+\frac z{2+zx+xy}\le$$ $$\le \frac x{(x+z)(x+y)}+\frac y{(y+z)(x+y)}+\frac z{(x+z)(z+y)}=$$ $$=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$

Roman83
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1 Answers1

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Hint:

$$(1-x)(1-y)(1-z)+(1-zx)\ge 0 \implies 2+ xy+yz \ge x+y+z+xyz$$

Macavity
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