\(2\le\sqrt{x}+\sqrt{4-x}\le2\sqrt{2}\) (1) (ĐK: \(\left\{{}\begin{matrix}x\ge0\\4-x\ge0\end{matrix}\right.\)\(\Leftrightarrow0\le x\le4\))
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}2\le\sqrt{x}+\sqrt{4-x}\\\sqrt{x}+\sqrt{4-x}\le2\sqrt{2}\end{matrix}\right.\) (\(0\le x\le4\))
\(\Leftrightarrow\left\{{}\begin{matrix}4\le4+2\sqrt{x\left(4-x\right)}\\4+2\sqrt{x\left(4-x\right)}\le8\end{matrix}\right.\) (\(0\le x\le4\))
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x\left(4-x\right)}\ge0\\\sqrt{x\left(4-x\right)}\le2\end{matrix}\right.\)(\(0\le x\le4\))
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(4-x\right)\le4\\0\le x\le4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\0\le x\le4\end{matrix}\right.\) (đpcm)