Maximize $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$ Given three non-negative (as pointed out by Calvin Lin) real numbers $x+y+z = 3$, find the maximum value of $\sqrt{2x + 13} + \sqrt[3]{3y+5} + \sqrt[4]{8z+12}$.
(Source : Singapore Math Olympiad 2012, Senior section, Round 1, question 29).
I tried using the fact that $2x +13, 8z + 12\ge 0$ to deduce that $y \le 11$, but I couldn't continue from there on. The answer should be an integer, since only integer answers were allowed.
The competition was designed for 15/16 year olds. A simple yet elegant solution would be nice.

For reference of the original problem, 

\begin{align*}&\sqrt{\dfrac{2x+13}{4}}\cdot\sqrt{4}+\sqrt[3]{\dfrac{3y+5}{4}}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}+\sqrt[4]{\dfrac{8z+12}{8}}\cdot\sqrt[4]{2}\cdot\sqrt[4]{2}\cdot\sqrt[4]{2}\\
&\le\dfrac{\dfrac{2x+13}{4}+4}{2}+\dfrac{\dfrac{3y+5}{4}+2+2}{3}+\dfrac{\dfrac{8z+12}{8}+2+2+2}{4}\\
&=\dfrac{1}{4}(x+y+z)+\dfrac{29}{4}\\
&=8
\end{align*}
