Đáp án:
$\frac{1}{1+x}+\frac{1}{1+y} \leq \frac{2}{1+\sqrt{xy}}$ với $x,y>0 , xy \leq1$
Giải thích các bước giải:
$\frac{1}{1+x}+\frac{1}{1+y} \leq \frac{2}{1+\sqrt{xy}}$
$→(\frac{1}{1+x}-\frac{1}{1 + \sqrt{xy}})+(\frac{1}{1+y}-\frac{1}{1+\sqrt{xy}})\leq0$
$→\frac{\sqrt{xy}-x}{(1+x)(1+\sqrt{xy})}$ $+$ $\frac{\sqrt{xy}-y}{(1+y)(1+\sqrt{xy})}$ $\leq 0$
$→\frac{\sqrt{y}-\sqrt{x}}{1 + \sqrt{xy}}(\frac{\sqrt{x}}{1 + x} -\frac{\sqrt{y}}{1+y}) \leq 0$
$→\frac{\sqrt{y}-\sqrt{x}}{1 + \sqrt{xy}}$ . $\frac{\sqrt{x}(1+y)-\sqrt{y}(1+x)}{(1+x)(1+y)}$ $\leq0$
$→\frac{\sqrt{y}-\sqrt{x}}{1 + \sqrt{xy}}$ . $\frac{\sqrt{x} + y\sqrt{x} -\sqrt{y} -x\sqrt{y} }{(1+x)(1+y)}$ $\leq0$
$→\frac{\sqrt{y}-\sqrt{x}}{1 + \sqrt{xy}}$ . $\frac{(\sqrt{xy} -1)(\sqrt{y} -\sqrt{x})}{(1+x)(1+y)}$ $\leq 0$
$→\frac{(\sqrt{y}-\sqrt{x})².(\sqrt{xy}-1)}{(1 + \sqrt{xy})(1+x)(1+y)}$ $\leq0$ luôn đúng $∀x,y>0 , xy \leq1$
Vậy $\frac{1}{1+x}+\frac{1}{1+y} \leq \frac{2}{1+\sqrt{xy}}$ với $x,y >0 , xy \leq1$
