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, 
 A: If $y$ is allowed to be negative, then there is no maximum.
Take $ x =k - \frac{13}{2}$, $ y = -k - \frac{5}{3}$, $ z= 3 + \frac{13}{2} + \frac{5}{3}$
The important part of the value looks like $ \sqrt{2k} + \sqrt[3]{-3k} + C$, which tends to infinity as $k\rightarrow \infty$.

Otherwise, if $y$ is positive, see Math110's solution.
A: \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*}
A: math110's AM-GM solution is no doubt (to me at least) what the author's wanted. But a less inspired approach is that with Lagrange Multipliers, you need to solve the system of equations:
$$\left\{\begin{aligned}
x+y+z&=3\\
\frac{1}{\sqrt{2x+13}}&=\lambda\\
\frac{1}{\sqrt[3]{(3y+5)^2}}&=\lambda\\
\frac{2}{\sqrt[4]{(8z+12)^3}}&=\lambda\\
\end{aligned}\right.$$
The second and third imply that $\sqrt{2x+13}=\sqrt[3]{(3y+5)^2}$, so $(2x+13)^3=(3y+5)^4$.
Similarly the second and fourth imply that $2\sqrt{2x+13}=\sqrt[4]{(8z+12)^3}$, so $16(2x+13)^2=(8z+12)^3$. 
Now you can forget about $\lambda$ and work with the system
$$\left\{\begin{aligned}
x+y+z&=3\\
(2x+13)^3&=(3y+5)^4\\
16(2x+13)^2&=(8z+12)^3\\
\end{aligned}\right.$$
Eliminating $z$:
$$\left\{\begin{aligned}
(2x+13)^3&=(3y+5)^4\\
(2x+13)^2&=4(9-2x-2y)^3\\
\end{aligned}\right.$$
If there are any solutions where the sides of these equations are integers, we would have to have some integer to the $12$th on both sides of the first equation. Trying something nice and small like $2^{12}$ implies that $x$ would have to be $1.5$ and $y$ would have to be $1$. And this is also a solution to the second equation. So one solution is $(x,y,z)=(1.5,1,0.5)$, which yields a value of $8$. Maybe the relative simplicity of the the curves in this last system can show that there are no other solutions: 
It's a little tedious, but we can solve for $y$ in each equation, and the difference of the two functions of $y$ is $$\frac{(2x+13)^{3/4}-5}{3}-\left(\frac{\left(\frac{(2x+13)^2}{4}\right)^{1/3}-9+2x}{2}\right)$$
whose derivative is 
$$\frac{1}{2(2x+13)^{1/4}}-\frac{2^{1/3}}{3(2x+13)^{1/3}}-1$$
For positve $x$, this is always negative. So the curves from the last system above can only cross once in the domain where $x$ is positive, and we already found where they do.
And then it remains to show that this yields a local maximum value, not a minimum value or a degenerate solution to the Lagrange multiplier problem. 
If there are constraints on $x$, $y$, $z$ being positive, then the boundaries of the planes $x=0$ (subject to $y+z=3$), $y=0$ (subject to $x+z=3$), and $z=0$ (subject to $x+y=3$) need to be separately checked, followed by the boundaries of those boundaries: $(3,0,0)$, $(0,3,0)$, and $(0,0,3)$.
