ĐKXĐ: \(x\ge0\)
Ta có:
\(VT=\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\le\sqrt{\frac{2}{x+3}+\frac{2}{3x+1}}=\sqrt{\frac{8\left(x+1\right)}{\left(x+3\right)\left(3x+1\right)}}=\frac{2}{\left(1+\sqrt{x}\right)}\sqrt{\frac{2\left(1+\sqrt{x}\right)^2\left(x+1\right)}{\left(x+3\right)\left(3x+1\right)}}\)
Mà \(\left(x+3\right)\left(3x+1\right)-2\left(1+\sqrt{x}\right)^2\left(x+1\right)=x^2-4x\sqrt{x}+6x-4\sqrt{x}+1=\left(\sqrt{x}-1\right)^4\ge0\)
\(\Rightarrow\left(x+3\right)\left(3x+1\right)\ge2\left(1+\sqrt{x}\right)^2\left(x+1\right)>0\)
\(\Rightarrow\frac{2\left(1+\sqrt{x}\right)^2\left(x+1\right)}{\left(x+3\right)\left(3x+1\right)}\le1\)
\(\Rightarrow VT\le\frac{2}{1+\sqrt{x}}\)
Dấu "=" xảy ra khi và chỉ khi \(x=1\)
