Find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$. 
How do I find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$?

I have tried plugging in values, tried manipulating and inequality by means but that didn't worked. How do I approach this question?
 A: $$\underbrace{(c-b)}_{\ge0}\underbrace{(b^2-a^2)}_{\ge0}+c^2\underbrace{(1-c)}_{\ge0}\\
$$
$b^2-a^2$ is the only term dependent on $a$ and thus it attains a maximum at $a=0.$ The problem then reduces to
$$b^2(c-b)+c^2(1-c)$$ for $0\le b\le c\le1.$ Sub $b=rc.$
$$r^2c^3(1-r)+c^2(1-c)\\
c^2(r^2(1-r)c+1-c)$$
$r^2(1-r)\overset{\partial_r}{\longrightarrow}r(2-3r)$ has a maximum at $r=\frac23.$
We then maximize
$$c^2\left(1-\frac{23}{27}c\right)\overset{\partial_c}{\longrightarrow}c\left(2-\frac{23}9c\right)\\
c=\frac{18}{23}\\
b=\frac23\cdot\frac{18}{23}=\frac{12}{23}$$
The maximum is thus $$\frac{108}{529}.$$
A: Capitalizing on writing the expression as a sum of two positive numbers, indicates the value of a:
$$\begin{align}E(a,b,c)&=\underbrace{(b^2-a^2)}_{\ge 0}\underbrace{(c-b)}_{\ge 0}+\underbrace{c^2(1-c)}_{\ge 0}, a,b,c \lt 1\\ &\le (b^2-0)(c-b)+c^2(1-c), \boxed{a= 0}\end{align}$$
The first term $b^2(c-b)$ could be analyzed using $f(x)=x^2(c-x), x\lt c$. This cubic has a maximum at $x=\frac 23 c$, therefore $\boxed{b=\frac 23 c}$
Replacing $b=\frac 23 c$, the original expression after considering $a=0$ becomes:
$$E(a,b,c)\lt \frac {4}{27}c^3+c^2(1-c), E(a,b,c)\lt -\frac{23}{27}c^3+c^2$$
The function $g(x)=-\frac{23}{27}x^3+x^2$ has a maximum at $x=\frac{18}{23}$. Therefore $\boxed{c=\frac{18}{23}}$.
Replacing $a,b,c$ finds the maximum of the original expression.
A: If you're interested in a quick answer, you can use quantifier elimination, for example in Mathematica:
$$\forall_{a,b,c,0\leq a\leq b\leq c\leq 1}\qquad a^2(b-c)+b^2(c-b)+c^2(1-c)<m$$
Then use Resolve[%, Reals] to obtain:
$$m>\frac{108}{529}.$$
Similarly, using Maximize, you can also obtain the values $a=0$, $b=\frac{12}{23}$, and $c=\frac{18}{23}$.
