If $x, y, z$ are positive real numbers with the property $ xy, yz, zx \leq 1 $, then prove that $$ \frac{3-(xy+yz+zx)}{2} \geq \sum_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2}.$$
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By that sum, you mean $\sum_{\text{cyc}}$ or $\sum_{\text{sym}}$ by any chance? – Ian Mateus Dec 28 '13 at 15:41
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It's cyclic sum! :D – Dec 28 '13 at 15:44
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Are you sure about the $y^x$, it makes it ugly... Bernoulli's inequality may be useful. – Ragnar Dec 28 '13 at 16:08
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1Hi, welcome to math.se. Please give more details on how you encountered the problem and any attempts you made on the problem by editting the question. It'd make it easier for others to help you. – Lost1 Dec 28 '13 at 16:15
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Oh no! There is $ y^2 $not $y^x$..Sorry – Dec 28 '13 at 16:15
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Use AM-GM to get: $$RHS = \sum_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2} \le \sum_{cyc}\frac{1-x^2y^2}{2+2xy}= \frac12\sum_{cyc}(1-xy) = LHS $$
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