Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$. I am working on inequality problems for mathematical olympiads. I have come across this problem:

Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $

I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.
I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.

Since $\frac{x}{10}, \frac{y}{10}, \frac{z}{10}\in[0; 1]$, we have that $\left(\frac{x}{10}\right)^2\leqslant\frac{x}{10}, \left(\frac{y}{10}\right)^2\leqslant\frac{y}{10}, \left(\frac{z}{10}\right)^2\leqslant\frac{z}{10}$. Thus \begin{align*}2x^2+y^2+z^2&=100\cdot \left(2\cdot\left(\frac{x}{10}\right)^2+\left(\frac{y}{10}\right)^2+\left(\frac{z}{10}\right)^2\right)\\&\leqslant 100\cdot \left(2\cdot \frac{x}{10}+\frac{y}{10} +\frac{z}{10}\right)\\&=100\cdot \left(\frac{x}{10}+1\right)\\&\leqslant 100\cdot (1+1)=200\end{align*} With equality if and only if $x=10, y=0, z=0.$
