Question

Chứng minh : \(0\le\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\le1\) \((x\ge0)\)

Answer 1

\(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge0\)(1)

\(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}-1=\dfrac{\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}=\dfrac{-\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}=-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le0\)

\(\Rightarrow\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\le1\) (2)

(1);(2) => đpcm