Find the maximum and minimum value of the expression $\frac{x+y+z}3+\frac{2016}{\sqrt[3]{xyz}}$ Find the maximum and minimum value of the expression 
$$\frac{x+y+z}3+\frac{2016}{\sqrt[3]{xyz}}$$
if $x,y,z \in [1,2016]$
My work so far:
Let $2016=a, x,y,z \in [1,a]$, $$\frac{x+y+z}3+\frac a{\sqrt[3]{xyz}}=A$$
$$m\le A \le M$$
Use $AM \ge GM:$
$$A=\frac{x+y+z}3+\frac a{\sqrt[3]{xyz}} \ge \sqrt[3]{xyz}+\frac a{\sqrt[3]{xyz}}\ge 2\sqrt a$$
So $A \ge m=2\sqrt a$, $(x=y=z=\sqrt a)$.
Help me find $M$.
 A: Well I'd just use calculus. We want to find the critical points of $f(x,y,z)=\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}$. So we want to solve the system
$$\begin{align*}
\frac{\partial f}{\partial x}=\frac{1}{3}-\frac{672\cdot yz}{xyz\cdot\sqrt[3]{xyz}}=0 \\
\frac{\partial f}{\partial y}=\frac{1}{3}-\frac{672\cdot xz}{xyz\cdot\sqrt[3]{xyz}}=0 \\
\frac{\partial f}{\partial z}=\frac{1}{3}-\frac{672\cdot yx}{xyz\cdot\sqrt[3]{xyz}}=0 \\
\end{align*}$$
Doing a little manipulation we can get the solution to be
$$x=y=z=\sqrt{2016}$$
Now to find the minimum and maximum of the expression we just compare the value of $f$ at the point we got as well as at the endpoints of the allowed domain. What we have is
$$\begin{align*}
f(1,1,1)=f(2016,2016,2016)=2017 \\
f(2016,1,2016)=f(2016,2016,1)=f(1, 2016, 2016)\approx 1356.97 \\
f(1,1,2016)=f(2016,1,1)=f(1,2016,1)\approx 832.25 \\
f(\sqrt{2016},\sqrt{2016},\sqrt{2016})\approx 89.8
\end{align*}$$
So as you can see the expression is at a minimum when $(x,y,z)=(\sqrt{2016},\sqrt{2016},\sqrt{2016})$ and it is at a maximum when $(x,y,z)=(1,1,1),(2016,2016,2016)$.
