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We are to maximise $x^{2}y-y^{2}x$, where $x,y \in [0,1]$. I've tried using AM-GM to find another (easier to maximise) expression, which gave me $xy(x-y) \le \frac{1}{2}(x^{2}+y^{2}) (x-y)$ but that doesn't appear to help.

The next part of the question asks to maximise $x^{2}y+y^{2}z+z^{2}x-x^{2}z-y^{2}x-z^{2}y$, where $x,y,z \in [0,1]$. I would imagine that this is simply triple the maximum for the first expression.

Any help would be appreciated.

NOTE: I shouldn't be using any calculus.

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2 Answers 2

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By AM-GM $$x^2y-xy^2=xy(x-y)\le x\cdot \left(\frac {y+(x-y)}2\right)^2=\frac{x^3}4\le \frac14$$ with equality at the first $\le$ iff $y=x-y$ (i.e. $x=2y$) and at the second $\le$ iff $x=1$. Hence $$x^2y-xy^2\le\frac 14$$ with equality iff $x=1$, $y=\frac 12$.

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Thank you! These questions always seem so easy once someone explains them. –  Daniel Littlewood Nov 11 '12 at 23:32

A small technicality here, AM-GM only works in general for positive reals, although in this case using it works because $ab\le\frac{(a+b)^2}{4}\Leftrightarrow\frac{(a-b)^2}{4}\ge 0$ is true for all reals.

As for the second question, it can be approached similarly: Note that $x^2y+y^2z+z^2x-xy^2-yz^2-zx^2=(x-z)(z-y)(y-x)$. Now two of $x-z, z-y, y-x$ have the same sign, so WLOG let them be $x-z, z-y$. Then we have $|x-z|+|z-y|=|x-y|$. Note that: $$(x-z)(z-y)(y-x)\le (|x-z||z-y|)|y-x|\le \frac{|x-y|^3}{4}\le \frac{1}{4}$$ Equality holds here when $y=1, x=0, z=\frac{1}{2}$, or the cyclic permutations.

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Thank you for covering the second part :-) –  Daniel Littlewood Nov 14 '12 at 21:56

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