Let's determine the maximum of the function $f(x)=\sqrt{x-2}+\sqrt{2 x-7}+\sqrt{18-3 x}$.
Solution. First, let's examine the domain of the function. The value under the square root cannot be negative, therefore $3.5 \leq x \leq 6$.

We apply the inequality between the arithmetic mean and the quadratic mean:

$$
\frac{a+b+c}{3} \leq \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}
$$

to the three square root expressions in the function:

$$
\frac{\sqrt{x-2}+\sqrt{2 x-7}+\sqrt{18-3 x}}{3} \leq \sqrt{\frac{(x-2)+(2 x-7)+(18-3 x)}{3}}=\sqrt{3}
$$

Thus,

$$
f(x)=\sqrt{x-2}+\sqrt{2 x-7}+\sqrt{18-3 x} \leq 3 \sqrt{3}
$$

Equality holds if and only if

$$
\sqrt{x-2}=\sqrt{2 x-7}=\sqrt{18-3 x} \Leftrightarrow x=5
$$

which is an element of the domain. Therefore, the maximum of the function is $3 \sqrt{3}$, which is attained (uniquely) at $x=5$.