$\frac{\sqrt{x}}{1+y}+\frac{\sqrt{y}}{1+x}=\frac{x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}}{xy+x+y+1}$Vì $0\leq x;y\leq 0,5$ $\Rightarrow \left ( \frac{1}{\sqrt{2}}-\sqrt{x}\right )\left ( \frac{1}{\sqrt{2}}-\sqrt{y}\right )\geq 0$
$\Rightarrow \sqrt{x}+\sqrt{y}\leq \frac{1}{\sqrt{2}}+\sqrt{2xy}$ (1)
Mặt khác $x\sqrt{x}+y\sqrt{y}\leq \frac{x+y}{\sqrt{2}}$ (2)
Dùng bđt Côsi cho 2 số ta có
$\sqrt{xy}\leq xy+\frac{1}{4}\Rightarrow \frac{2\sqrt{2xy}}{3}\leq \frac{2\sqrt{2}xy}{3}+\frac{\sqrt{2}}{6}$ (3)
$\sqrt{xy}\leq \frac{x+y}{2}\Rightarrow \frac{\sqrt{2xy}}{3}\leq \frac{\sqrt{2}(x+y)}{6}$(4)
Lấy (3)+(4) ta có $\sqrt{2xy}\leq \frac{2\sqrt{2}xy}{3}+\frac{\sqrt{2}\left ( x+y \right )}{6}+\frac{\sqrt{2}}{6}$ (5)
Thay (5) vào (1) rồi cộng với(2) ta có $x\sqrt{x}+y\sqrt{y}+\sqrt{x}+\sqrt{y}\leq \frac{2\sqrt{2}}{3}\left ( xy+x+y+1 \right )$
Ta có đpcm