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.

Some comments:
Both Tavish and Calvin Lin's solutions are good and are not just lucky. Let's break it down:
Lemma: If $a\ge 0, b\ge 0$ and $a+b$ is fixed, then $a^2+b^2$ obtains its maximum when either $a=0$ or $b=0$.
This can be proved either via property of parabola, or just algebraically. But it's easy to write it simply as $a^2+b^2 \le (a+b)^2$, and I believe nobody is against this.
Now if we fix $y+z$ we have $$2x^2 + (y^2 + z^2) \le 2x^2+ (y+z)^2 = 2x^2+ (10-x)^2 \tag 1$$
Next we apply the lemma on $x$ and $10-x$:
$$x^2+\left( x^2+(10-x)^2 \right)\le x^2 + (x+10-x)^2 = x^2 + 100 \tag 2$$
$$\le 100+100.\blacksquare$$
You can think of Calvin Lin's way as combining these two steps (or a generalized version of the lemma with more than two variables):
$$2x^2+y^2+z^2 = x^2 + (x^2+y^2+z^2) \le x^2 + (x+y+z)^2 = x^2 + 10^2 \le 200$$
Tavish's way is to apply the parabola property twice.

It's best understood from the picture below:

