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I want to find the the supremum of $\dfrac{|x|^{2/3}-|y|^{2/3}}{|x-y|^{2/3}} $ in the unit ball centered at the origin . Here $x\neq y$, $x,y \in \mathbb R^n$. How do I proceed ? Thank you for your help.

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Hint: By a modification of the triangular inequality you see that $$||x|^{2/3}-|y|^{2/3}|\leq |x-y|^{2/3}$$ for all $x\neq y$. So we have$$\dfrac{|x|^{2/3}-|y|^{2/3}}{|x-y|^{2/3}}\leq 1$$ Is the value of the function $=1$ at some point?

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Yup at $x=0, y=1$ . – Theorem Jul 13 '12 at 10:51
So does this answer your question? – Simon Markett Jul 13 '12 at 10:59
+1 for the maieutics. – Did Jul 13 '12 at 11:31
@SimonMarkett : Yup :) . thanks a ton . – Theorem Jul 13 '12 at 11:41
Btw it's $x=1$, $y=0$. @did I had to look up maieutics...but thx ;) – Simon Markett Jul 13 '12 at 11:48

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