9 The maximum value of the function $f(x)=2 \sqrt{x-3}+\sqrt{5-x}$ is
9 $\sqrt{10}$ Hint: By Cauchy's inequality, we have
$$
(2 \sqrt{x-3}+\sqrt{5-x})^{2} \leqslant\left(2^{2}+1^{2}\right)(x-3+5-x)=10,
$$

with equality holding if and only if $\frac{\sqrt{x-3}}{2}=\frac{\sqrt{5-x}}{1}$, i.e., when $x=\frac{23}{5}$.
Therefore, the maximum value of $f(x)=2 \sqrt{x-3}+\sqrt{5-x}$ is $\sqrt{10}$.