8. Let $a, b, c$ be positive real numbers such that $a+b+c=2013$. Find the maximum value of $\sqrt{3 a+12}+\sqrt{3 b+12}+\sqrt{3 c+12}$.
8. Answer: 135
Solution. Note that
$$
\begin{aligned}
(\sqrt{3 a+12} & +\sqrt{3 b+12}+\sqrt{3 c+12})^{2} \leq 3(3 a+12+3 b+12+3 c+12) \\
& =9(a+b+c+12)=9(2013+12)=9 \times 2025,
\end{aligned}
$$
where the equality holds if $3 a+12=3 b+12=3 c+12$, i.e., $a=b=c=671$. Thus the answer is $3 \times 45=135$.