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hotien217

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huynhbachkhoa23

Trường hợp $x \ge y$
$x^2+y^2=(x-y)x+y(x+y) \ge (\sqrt{2013}+\sqrt{2014}-\sqrt{2015})x+\sqrt{2015}(x+y) \ge ...$
Trường hợp $x\le y$ thì $y\ge x \ge \sqrt{2013}+\sqrt{2014}$
$x^2+y^2 \ge 2(\sqrt{2013}+\sqrt{2014})^2$
 
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