find numbers x,y,z that achieve maximum possible value of $E(x,y,z) = \alpha(x-y)^2 + \beta(y-z)^2 + \gamma (z-x)^2$ 
Let $\alpha, \beta, \gamma$ be positive real numbers and let $[a,b]$ be an interval. Find numbers $x,y,z\in [a,b]$ so that $E(x,y,z) =  \alpha(x-y)^2 + \beta(y-z)^2 + \gamma (z-x)^2$ is maximum.

I think we can assume WLOG that $\alpha \leq \beta\leq \gamma.$

But how would one justify the above assumption?

Also, I think one can assume $x\leq y\leq z$.

As for the justification, under the assumption that $x\leq y\leq z,$ one can see that $E(x,y,z) \ge E(y,x,z)$ and $E(x,y,z) \ge E(z,y,x).$ But $E(x,y,z) - E(z,y,x) = (\alpha - \beta)((x-y)^2 - (y-z)^2),$ and it's not clear that this value is nonnegative (if one only makes the above two additional assumptions).

With the above two assumptions, note that a convex function on a bounded interval attains its maximum at one of the endpoints. It might be useful to define a function like $f_1(y) = \alpha(y-a)^2 + \beta(y-b)^2,$ which can be used to maximize $E(a,y,b)$ where $y$ ranges over $[a,b]$. But must the maximum value of $E$ be obtained at a point of the form $(a,y,b)$ or $(b,y,a)$?
 A: The maximum is attained with two of $x, y, z$ at the endpoints $a, b$. Because if one or both of the endpoints are not used, suppose $x \le y \le z$ (same proof with other orders), then
$\alpha (x-y)^2 \le \alpha(a-y)^2$,
$\beta (y-z)^2 \le \beta(y-b)^2$,
$\gamma(z-x)^2 \lt \gamma(b-a)^2$,
so $E(x,y,z) \lt E(a,y,b)$.
We assume w.l.o.g. that $\alpha \le \beta \le \gamma$.
We compare $E$ of all permutations of $x,y,z$. Note that permuting $(x,y,z)$ is equivalent to permuting $((x-y)^2, (y-z)^2, (z-x)^2)$.
It is a known result that the maximum sum of term-by-term product of two finite sequences is when they are ordered the same way (Proof for two sequences producing the maximum value when sorted).
So $\alpha \le \beta \le \gamma$ implies $(x-y)^2 \le (y-z)^2 \le (z-x)^2$ for the maximum.
As two of $x,y,z$ must be $a,b$, this implies $\{x,z\}=\{a,b\}$.
We can choose $x=a, z=b$ as we obtain the same value for $E$ by symmetry: $E(a,y,b)=E(b,a+b-y,a)$.
Then to get the optimum value for $y$, study $\frac {\partial E(x,y,z)} {\partial y} = 2\alpha(y-x)+2\beta(y-z) = 2\alpha(y-a)+2\beta(y-b)$.
This is null for $y_0$ such that $\alpha(y_0-a)=\beta(b-y_0)$, $y_0=\frac {\alpha a + \beta b} {\alpha+\beta}$.
$\frac {\partial^2 E(x,y,z)} {\partial y^2} = 2\alpha+2\beta > 0$, so $\frac {\partial E(x,y,z)} {\partial y}$ is a strictly increasing function of $y$ that is null on $y_0$.
$E(a,y,b)$ is a decreasing then increasing function of $y$, that reaches a minimum on $y_0$.
There are 2 local maxima, one on $a$ and one on $b$. The largest one is for $y=a$:
$E(a,a,b)-E(a,b,b)$
$=\alpha(a-a)^2+\beta(a-b)^2+\gamma(b-a)^2-\alpha(a-b)^2-\beta(b-b)^2-\gamma(b-a)^2$
$=(\beta-\alpha)(a-b)^2 \ge 0$.
In summary, maximum is attained at $x=a,y=a,z=b$.
(I don't detail the other solutions. There is the symmetric one at $(b,b,a)$; plus some other permutations of $(a,a,b)$ and $(a,b,b)$ when there are some equalities between $\alpha, \beta, \gamma$. Those other solutions have the same maximum value of $E$).
A: Here is a solution based on a geometrical interpretation. One can object that it is not fully rigorous, but its interest is to give a "big picture" of this issue.
First of all, we are concerned by the range of function $E$, i.e., the set of values of $m$ such that
$$E(x,y,z)=\alpha(x-y)^2 + \beta(y-z)^2 + \gamma (z-x)^2=m\tag{1}$$
has a solution $(x,y,z)$. We are going to see that it is an interval: $m \in (0,M)$, the "restricted issue" being to find for which $(x,y,z)$ one obtains extremal value $M$.
First of all, equation (1) is that of a cylinder $(C_m)$ with axis directed by $(1,1,1)$. Indeed, this is the equation of a quadric surface invariant by any translation:
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\to \begin{pmatrix}x\\y\\z\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}$$
(i.e., if we replace $x,y,z$ by $x+t,y+t,z+t$ resp., equation (1) is left unchanged)
Remark: Cylinder $(C_m)$ belongs to the family of elliptical cylinders (the other categories, hyperbolic or parabolic cylinders being ruled out due to the fact that $E(x,y,z)$ is a positive definite expression). An elliptical cylinder has an elliptical normal cross section.
The constraint on $x,y,z$ has, as well, a geometrical interpretation
$$(x,y,z) \in \underbrace{[a,b]^3}_{(B)}$$
where $(B)$ is a cube. One of the great diagonals of $(B)$, joining vertices $(a,a,a)$ and $(b,b,b)$, is also directed by vector $(1,1,1)$.
The constraint is that cube $(B)$ must meet $(C_m)$.
As cylinder $(C_m)$ is "inflating" when $m$ increases, it intersects the cube till a certain limit value of $m$ which is reached when the "last" vertex of cube $(B)$ among
$$(a,a,b),(a,b,a),(b,a,a),(b,b,a),(b,a,b),(a,b,b)$$
is just "touching" the cylinder. Please note that we don't consider $(a,a,a)$ and $(b,b,b)$ : they "stay in place" on the diagonal directed by vector $(1,1,1)$.
Moreover, this issue has a symmetry due to the fact that the center of the cube:
$$\Omega\left(\frac{a+b}{2},\frac{a+b}{2},\frac{a+b}{2}\right)$$ is a center of symmetry for the cube and for the cylinder. As
$$\begin{cases}(a,a,b) \ \text{and} \ (b,b,a)\\
(a,a,b) \ \text{and} \ (b,b,a)\\
(a,a,b) \ \text{and} \ (b,b,a)\end{cases} resp.$$
are symmetrical with respect to $\Omega$ (said otherwise, their midpoint is $\Omega$), we need only consider 3 vertices, for example:
$$\{(a,a,b),(a,b,a),(b,a,a)\}\tag{2}$$
Final answer: considering (2), as
$$\begin{cases}E(a,a,b)&=&(\beta+\gamma)(a-b)^2\\E(a,b,a)&=&(\alpha+\beta)(a-b)^2\\E(b,a,a)&=&(\alpha+\gamma)(a-b)^2\end{cases}$$
the maximum is reached in the following vertices:
$$\begin{cases}
(a,a,b) \ \& \ (b,b,a) \ \ \text{if} \ \ \beta+\gamma \ge \alpha+\beta, \alpha+\gamma  \ \ \iff \ \ \alpha \le \beta,\gamma\\
(a,b,a) \ \& \ (b,a,b) \ \ \text{if} \ \ \alpha+\beta \ge \beta+\gamma,\alpha+\gamma  \ \ \iff \ \ \gamma \le \alpha,\beta\\
(b,a,a)  \ \& \ (a,b,b)\ \ \text{if} \ \ \alpha+\gamma \ge \alpha+\beta,\beta+\gamma   \ \ \iff \ \ \beta \le \alpha,\gamma \end{cases}$$
(with the meaning $p \le q,r \iff p \le q \ \ \& \ \ p 
 \le r$)

A remark: Quadratic form $E(x,y,z)$ has
$$Q=\begin{pmatrix}\alpha+\gamma&-\alpha&-\gamma\\
-\alpha&\alpha+\beta&-\beta\\
-\gamma&-\beta&\beta+\gamma\\
\end{pmatrix}$$
as its associated matrix ; in particular vector $\begin{pmatrix}1\\1\\1
\end{pmatrix}$  can be interpreted as an eigenvector of $Q$ associated with eigenvalue $0$, otherwise said, a basis of the kernel of $Q$.
But beyond that, using linear algebra doesn't look
the right approach.
