Function satisfiyng Hölder condition but not Lipschitz condition.

Let $f:[0,+\infty)\to[0,+\infty)$ be defined by $f(x)=x^{1/n}$ where $n\in\mathbb{N}$. Show that $f$ satisfies $|f(x)-f(y)|\leq 2^{(n-1)/n}|x-y|^{1/n}$. Prove that $f$ isn't Lipschitz in any interval cointaining $0$.

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What have you tried? –  Alex Becker May 12 '12 at 2:29
Dear Gastón, in order to get some help for questions, it could be good to show what you have tried so far and, in the future, try to write questions not in an "exercise" fashion, but showing what was your effort so far. –  matgaio May 12 '12 at 2:29
I dont know how to prove the first inequallity for Hölder condition, i dont know many useful inequallities. The last part its ok. –  Gastón Burrull May 12 '12 at 3:27
The inequality is here: math.stackexchange.com/questions/144434/… –  Gastón Burrull May 13 '12 at 2:24
Note that "$f'(0)=+\infty$".
Formally, given $\epsilon>0$, you can find $\delta>0$ such that $x\in[0,0+\delta)$ implies $|f(x)-f(0)|>\epsilon|x-0|$, and in particular, $f$ cannot be Lipschits