Cho $a,b,c >0$ thõa mãn $abc=1$
CMR : $\sum {\sqrt{\dfrac{2}{a^2+1}}} \leq 3$

Giả sử $a=max\left \{ a,b,c \right \}\Rightarrow bc\leq 1$
Ta đi chứng minh $\dfrac{1}{\sqrt{1+b^{2}}}+\dfrac{1}{\sqrt{1+c^{2}}}\leq \dfrac{2}{\sqrt{1+bc}}$
  Ta có :   $\dfrac{1}{2}\left ( \dfrac{1}{\sqrt{1+b^{2}}} +\dfrac{1}{\sqrt{1+c^{2}}}\right )^{2}\leq \dfrac{1}{1+b^{2}}+\dfrac{1}{1+c^{2}}=\dfrac{1-b^{2}c^{2}}{\left ( 1+b^{2} \right )\left ( 1+c^{2} \right )}+1\leq 1+\dfrac{1-b^{2}c^{2}}{\left ( 1+bc \right )^{2}}=\dfrac{2}{1+bc}$
  Lại có :   $\dfrac{1}{\sqrt{1+a^{2}}}\leq \dfrac{\sqrt{2}}{1+a}$
Ta chỉ cần chứng minh : $\dfrac{1}{1+a} +\dfrac{\sqrt{2}}{\sqrt{1+bc}}\leq \dfrac{3}{2}\Leftrightarrow \dfrac{\left ( \sqrt{1+a}-\sqrt{2a}\right )^{2}}{2\left ( 1+a \right )}\geq 0$ (luôn đúng )
=> ĐPCM