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.

Let $u=\sqrt2x$, $y=v$, $z=w$. Then, it is equivalent to find the maximum $k^2=u^2+v^2+w^2$ subject to
$$\frac u{10\sqrt2} + \frac v{10} + \frac w{10} =1\tag1$$
Observe that $k$ is the distance from the origin to the plane (1) with the axis-intercepts $10\sqrt2,10,10$. In the first quadrant, (1) is a tetrahedron with its furthest corner at the $x$-intercept $10\sqrt2$. Thus, the maximum is $k^2_{max}=(10\sqrt2)^2 =200$.
