Áp dụng \(a+b\le\sqrt{2\left(a^2+b^2\right)}\)
\(f\left(x\right)\le\sqrt{2\left(x+3+6-x\right)}=3\sqrt{2}\)
Dấu "=" xảy ra khi \(x+3=6-x\Leftrightarrow x=\frac{3}{2}\)
Áp dụng \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
\(f\left(x\right)\ge\sqrt{x+3+6-x}=3\)
Dấu "="x ảy ra khi \(\left[{}\begin{matrix}x+3=0\\6-x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)