ĐKXĐ: \(x^2\ge2\)
Đặt \(\sqrt{x^2-2}=t\ge0\)
\(\Leftrightarrow\frac{1}{\sqrt{t^2+3}}+\frac{1}{\sqrt{3t^2+1}}\le\frac{2}{t+1}\)
Ta có: \(\frac{1}{\sqrt{t^2+3}}+\frac{1}{\sqrt{3t^2+1}}\le\sqrt{2\left(\frac{1}{t^2+3}+\frac{1}{3t^2+1}\right)}=2\sqrt{\frac{2\left(t^2+1\right)}{\left(t^2+3\right)\left(3t^2+1\right)}}\) (1)
Mặt khác ta luôn có:
\(\left(t-1\right)^4\ge0\Leftrightarrow t^4-4t^3+6t^2-4t+1\ge0\)
\(\Leftrightarrow3t^4+10t^2+3\ge2t^4+4t^3+4t^2+4t+2\)
\(\Leftrightarrow\left(t^2+3\right)\left(3t^2+1\right)\ge2\left(t+1\right)^2\left(t^2+1\right)\)
\(\Leftrightarrow\frac{2\left(t^2+1\right)}{\left(t^2+3\right)\left(3t^2+1\right)}\le\frac{1}{\left(1+t\right)^2}\) (2)
(1);(2) \(\Rightarrow VT\le2\sqrt{\frac{1}{\left(1+t\right)^2}}=\frac{2}{t+1}=VP\)
\(\Rightarrow\) BPT đã cho luôn đúng với mọi \(t\) hay nghiệm của BPT là \(x^2\ge2\)
