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.
 A: 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:

A: 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 $k^2_{max}=(10\sqrt2)^2 =200$.
A: Substitute the value of $z$ in the expression: $$2x^2 +y^2 +(x+y-10)^2 $$ As increasing $x$ and $y$ increases the value of this expression, the maximum occurs when $x+y$ is maximum, i.e. $x+y=10$: $$2x^2 +(10-x)^2 = 3x^2 -20x +100$$
This is a parabola which attains its maximum value on $[0,10]$ at $x=10$, which is $200$.
A: For a calculus-based method that delivers both solutions, you want a Lagrange multiplier. Let: \begin{align*}f\left(x,y,z\right) &= 2x^2 + y^2 + z^2 \\ g\left(x,y,z\right) &= x + y + z - 10\end{align*} The condition for an extremum is $$\vec{\nabla} f\left(x,y,z\right) = \lambda \vec{\nabla} g\left(x,y,z\right)$$ Calculating this out, we have a system with four equations and four unknowns: \begin{align*}4x &= \lambda \\ 2y &= \lambda \\ 2z &= \lambda \\ 0 &= x+y+z-10\end{align*}
Solving for $\lambda$ first is easiest: $$0 = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{2} - 10 \hspace{2.00cm} \rightarrow \hspace{2.00cm} \lambda = 8$$ With $\lambda=8$ figured out, we know one extreme point to be: $$x = 2  \hspace{2.54cm} y = 4 \hspace{2.54cm} z = 4$$

So what about the maximum? The best hint, which undoubtedly drives the other solutions to this, is the factor of $2$ in front of $x^2$, meaning this term grows faster than its counterparts. It only makes sense to try $y=z=0$, so that $x=10$ is evident from inspection. This makes the whole $\lambda$-apparatus somewhat useless, but at least the form is there in case the problem was trickier than it looked.
A: 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 $x=10, y=0, z=0.$
