Finding Maximum Under Constraint Suppose $a$,$b$,$c$ satisfy $a+b+c=1$ and $a$,$b$,$c\in [0,1]$
Find the maximum value of $(a-b)(b-c)(c-a)$
 A: WOLOG, suppose $c$ is the largest. If $a>b$, then $(a-b)(b-c)(c-a)\le 0$. So we may suppose $a<b<c$. Let $b-a=x\ge 0, c-b=y\ge 0$, then $3a+2x+y=1$, thus $2x+y\le 1$. The problem becomes find maximum value of 
$$
xy(x+y), \quad x, y \ge0.
$$ 
So we may suppose $2x+y=1$, then
$$
xy(x+y)=x(1-2x)(1-x)=:f(x), \quad 0\le x\le \frac{1}{2}.
$$
Put $f'(x)=1-6x+6x^2=0$, when $x=\frac{3-\sqrt{3}}{6}$, $f(x)$ attains the maximum.
A: Let us rename $a,b,c$ as $x,y,z$. From the constraint $x+y+z=1$ we may express $z$ as $z=1-x-y$ and study the function
$$
f(x,y) = (x-y)(y-z)(z-x) = (x-y)(2y+x-1)(1-2x-y)
$$
in the triangle
$$
T=\{(x,y)\in\mathbb R^2 : x,y\geq 0, ~x+y\leq1\}
$$
The partial derivatives of $f$ are
$$
\begin{cases}
f_x = -6x^2 - 6xy + 3y^2 + 6x -1\\
f_y = -3x^2 + 6xy - 6y^2 - 6y +1
\end{cases}
$$
The critical points of $f$ are the solutions of
$$
\begin{cases}
-6x^2 - 6xy + 3y^2 + 6x - 1 & = 0\\
-3x^2 + 6xy - 6y^2 - 6y + 1 & = 0
\end{cases}
~\Leftrightarrow~
\begin{cases}
-6x^2 - 6xy + 3y^2 + 6x - 1 & = 0\\
3x^2 + y^2 - 2x + 2y & = 0
\end{cases}
$$
from which you have $y=-1\pm\sqrt{1-3x^2+2x}$. Since $0\leq y\leq 1$, no solution $(x,y)$ can be inside $T$, therefore in $T$ $f$ does not have any critical point.
It follows that the maximum of $f$ is achieved on the boundary of $T$, i.e. either when $x=0$, $y=0$ or $y=1-x$:
1. $x=0$
$$
f(0,y) = y(y-1)(2y-1)
$$
when $y$ ranges in $[0,1]$ has maximum in $y=\frac{3-\sqrt3}6$, and
$$
f(0,\textstyle{\frac{3-\sqrt3}6}) ~=~ \frac{\sqrt3}{18}
$$
2. $x=0$
$$
f(x,0) = x(x-1)(1-2x)
$$
when $x$ ranges in $[0,1]$ has maximum in $x=\frac{3+\sqrt3}6$, and
$$
f(\textstyle{\frac{3+\sqrt3}6},0) ~=~ \frac{\sqrt3}{18}
$$
3. $y=1-x$
$$
f(x,1-x) = x(x-1)(2x-1)
$$
is analogous to case 1. w.r. to variable $y$.
Therefore we can conclude that the maximum is achieved when either $(a,b,c)$, $(b,c,a)$ or $(c,a,b)$ is $\big(\textstyle{\frac{3+\sqrt3}6},\textstyle{\frac{3-\sqrt3}6},0\big)$. In any case the maximum is
$$
(a-b)(b-c)(c-a) = \frac{\sqrt 3}{18}
$$
Edit (2): corrected 2 mistakes.
A: Another way is not to use calculus. WLOG, let $a=$MAX{$a,b,c$},then if we want max, it must >0, $→ a>c>b$, rewrite it to $\left(a-b\right)\left(c-b\right)\left(a-c\right)$, since $b\ge0$, so $0<a-b\le a,0<c-b\le c → \left(a-b\right)\left(c-b\right)\left(a-c\right)\le ac\left(a-c\right)$, then $b=0$ hold max,since $b=0$ will make $a,c$ much bigger also which make ac(a-c) bigger also, so we can sure $b=0$ is the condition ,ie, $a+c=1$ ,let $a=\cos ^2 x,c=\sin ^2 x$, we have:
$ac\left(a-c\right)=\left(\cos x *\sin x\right)^2\left(\cos ^2x-\sin ^2x\right)=\dfrac{\sin ^2\left( 2x\right)* \cos 2x}{4}=\dfrac{\left(1-\cos \left(4x\right)\right)\cos \left(2x\right)}{8}=\dfrac{\sqrt{\dfrac{\left(1+\cos 4x\right)\left(1-\cos 4x\right)^2}{2}}}{8}=\dfrac{\sqrt{\left(2+2\cos 4x\right)\left(1-\cos 4x\right)\left(1-\cos 4x\right)}}{16}\le \dfrac{\sqrt{\left(\dfrac{4}{3}\right)^3}}{16}=\dfrac{\sqrt{3}}{18}$ (AM-GM), the max will be hold when $1-\cos 4x=2+2coos4x$, ie.
$\cos 4x=-\dfrac{1}{3}→ \cos 2x=\sqrt{\dfrac{1-1/3}{2}}=\dfrac{\sqrt{3}}{3} → \cos ^2x=\dfrac{1+\dfrac{\sqrt{3}}{3}}{2}=\dfrac{3+\sqrt{3}}{6}=a, c=\dfrac{3-\sqrt{3}}{6}$
