ĐKXĐ: \(0\le x\le1\)
Do \(0\le x\le1\Rightarrow\left(1+2\sqrt{1-x}\right)-\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x}\left(2-\sqrt{1+x}\right)\ge0\)
\(\Rightarrow1+2\sqrt{1-x}\ge1+\sqrt{1-x^2}>0\)
\(\Rightarrow VP=\frac{1+2\sqrt{1-x}}{1+\sqrt{1-x^2}}\ge1\)
Mặt khác \(\frac{x}{x^2+3}-\frac{1}{4}=\frac{4x-x^2-3}{4\left(x^2+3\right)}=\frac{\left(x-1\right)\left(3-x\right)}{4\left(x^2+3\right)}\le0\) \(\forall x\in\left[0;1\right]\)
\(\Rightarrow\frac{x}{x^2+3}\le\frac{1}{4}\Rightarrow VT=2\sqrt{\frac{x}{x^2+3}}\le1\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi \(x=1\)
