<s>[INST] 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?
 [/INST] 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.
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