<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] $$\underbrace{(c-b)}_{\ge0}\underbrace{(b^2-a^2)}_{\ge0}+c^2\underbrace{(1-c)}_{\ge0}\\
$$
$b^2-a^2$ 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 $$\frac{108}{529}.$$
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