find the maximum of sum of roots without calculus A question asks me to evaluate the maximum and minimum value of the following function:
$$y=\sqrt{x}+\sqrt{27+x}+\sqrt{13-x}, 0\le x\le13$$
I have tried to compute y using $$x=0,1,2,...,13$$ I believe such maximum occurs at x=9, and the minimum value lies on boundary points, but how can I prove that without calculus?
Thanks!

Shifting the OP's function $9$ to the left we get a function of the form
$$g(x)=\sqrt{x+(n+1)^2}+\sqrt{x+\bigl(n(n+1)\bigr)^2}+\sqrt{n^2-x}.$$
Now it's easy to prove without calculus simply by squaring that $\sqrt{x+a^2}\leq \frac{1}{2a}x+a$ and $\sqrt{a^2-x}\leq-\frac{1}{2a}x+a$ for positive $a$ and $-a^2< x<a^2$.  Equality occurs iff $x=0$.  From here
$$\sqrt{x+(n+1)^2}\leq\frac{1}{2(n+1)}x+n+1,$$
 $$\sqrt{x+\bigl(n(n+1)\bigr)^2}\leq\frac{1}{2n(n+1)}x+n(n+1),$$
$$\sqrt{n^2-x}\leq-\frac{1}{2n}x+n.$$
Now adding up yields $g(x)\leq n+1+n(n+1)+n$ with equality iff $x=0$.
Hence the OP's function (with $n=2$) has its global maximum in $x=9$ of value $11$.
