Đặt $(x;y;z)=(\frac{b}{a};\frac{c}{b};\frac{a}{c})\rightarrow VT=\Sigma \frac{1}{\sqrt{x^2+1}}$Xài: $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\leq \frac{2}{\sqrt{1+ab}}$ với $ab<1$
G/s: $z\geq 1\rightarrow VT\leq \frac{1}{\sqrt{z^2+1}}+\frac{2}{1+xy}$ $($do $xyz=1)$
Đặt $t=1/z$
$\rightarrow VT\leq \frac{\sqrt{2}t}{1+t}+\frac{2}{\sqrt{1+t}}=\frac{\sqrt{2}t+2\sqrt{1+t}}{t+1}$
$\rightarrow $ cần c/m:$.......\leq \frac{3}{\sqrt{2}}\Leftrightarrow 2t+2\sqrt{2(t+1)}\leq 3t+3$