\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}< \dfrac{1}{3}\left(x\ge0;x\ne1\right)\\ \Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}< 0\\ \Leftrightarrow\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\dfrac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\left(\circledast\right)\)
Với \(x\ge0;x\ne1\) thì \(\left\{{}\begin{matrix}-\left(\sqrt{x}-1\right)^2< 0\\3\left(x+\sqrt{x}+1\right)>0\end{matrix}\right.\)
Vậy BĐT (*) được chứng minh luôn đúng với số x không âm khác -1, hay BĐT đầu luôn đúng.
\(< =>\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}< 0\)
\(< =>\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(< =>\dfrac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}\right)}< 0\)
\(< =>\dfrac{-\left(\sqrt{x}-1\right)^2}{3\left(\sqrt{x}+\dfrac{1}{2}\right)+\dfrac{9}{4}}< 0\left(\forall x\ge0,x\ne1\right)\)(luôn đúng)
