Hölder condition for the function $f(x)=x^{\frac{1}{n}}$

Question

I think my algebra is not very good.. I'm trying to show that $f:[0,\infty)\to[0,\infty)$ defined as $f(x)=x^{\frac{1}{n}}$ ,$n\in\mathbb{N}$ satisfies the Hölder condition $|f(x)-f(y)|\leq c|x-y|^{\frac{1}{n}}$ with $c=2^{\frac{n-1}{n}}$. Any hint will be appreciated...

Answer 1

$$(x+y)^{\alpha }=\frac{x+y}{(x+y)^{1-\alpha}}\leq \frac{x}{x^{1-\alpha}}+\frac{y}{y^{1-\alpha}}=x^{\alpha} +y^{\alpha}$$ for $x,y\geq 0 ,0\leq \alpha \leq 1.$ Hence by symmetry of $x$ and $y$ we get $$|(x+y)^{\alpha } - y^{\alpha } |\leq x^{\alpha}.$$ Now putting $x=u-v, y=v , $ where $u\geq v $ we obtain $$|u^{\alpha} - v^{\alpha} |\leq (u-v)^{\alpha }$$ and consequently $$|u^{\alpha} - v^{\alpha} |\leq |u-v|^{\alpha }.$$ So now you can take $\alpha=\frac{1}{n}.$