Given $x;y;z$ satisfying $0 \le x \le y \le z \le 1 $ Find the maximum value $$P=x^{2}(y-z)+y^{2}(z-y)+z^{2}(1-z)$$ $\bullet$ Note that: $x^2(y-z) \le 0$
$\bullet$ By AM-GM we have: $y^2(z-y)=\dfrac{y^2(2z-2y)}{2}\le\bigg(\dfrac{2y+2x-2y}{3} \bigg)^3 . \dfrac{1}{2}=\dfrac{4z^3}{27}$
$\bullet$ So we need to find the maximum value of $\dfrac{4z^3}{27}+z^2(1-z)$, I have found it, $P\le \dfrac{108}{529}$ when $x=0;y=\dfrac{12}{23};z=\dfrac{18}{23}$ and by Wolfram Alpha, the problem has been solved.
But I found the solution because I already know its maximum value, which is $P\le \dfrac{108}{529}$ when $x=0;y=\dfrac{12}{23};z=\dfrac{18}{23}$, is there any way without knowing the answer in advance to solve this problem ?
You don't need WA.
Just by AM-GM twice we obtain: $$x^2(y-z)+y^2(z-y)+z^2(1-z)\leq4\cdot\left(\frac{y}{2}\right)^2(z-y)+z^2(1-z)\leq$$ $$\leq4\left(\frac{2\cdot\frac{y}{2}+z-y}{3}\right)^3+z^2(1-z)=\frac{4z^3}{27}+z^2(1-z)=$$ $$=\frac{1}{27}z^2(27-23z)=\frac{4}{27\cdot529}\left(\frac{23z}{2}\right)^2(27-23z)\leq$$ $$\leq\frac{4}{27\cdot529}\left(\frac{2\cdot\frac{23z}{2}+27-23z}{3}\right)^3=\frac{108}{529}.$$ The equality occurs for $x=0$, $\frac{y}{2}=z-y$ and $\frac{23z}{2}=27-23z,$ which says that we got a maximal value.