Inequality between norms in $\mathbb{R}^n$ I am trying to prove that given $p>1$ there exists a constant $C=C(p,n)$ such that $\big||x|^px-|y|^py\big|\leq C\big(|x|^p+|y|^p\big)|x-y|$ for all $x,y\in\mathbb{R}^n$. It seems useful to consider the inequality $|x+y|^p\leq C\big(|x|^p+|y|^p\big)$ but I don´t know how to follow.
Any help will be appreciated!
 A: Let $f(x)=|x|^p\,x$. Then $f$ has a continuous derivative given by $f'(x)=(p+1)\,|x|^p$. By the mean value theorem, there exists a point $\xi$ between $x$ and $y$ such that
$$
|x|^p\,x-|y|^p\,y=(p+1)\,|\xi|^p\,(x-y).
$$
Since $\xi$ is between $x$ and $y$
$$
|\xi|^p\le\max(|x|^p,|y|^p)\le|x|^p+|y|^p.
$$
Then
$$
\bigl|\,|x|^p\,x-|y|^p\,y\,\bigr|\le(p+1)\,(|x|^p+|y|^p)\,|x-y|.
$$
Note that the argument is valid for $p>0$.
A: A different approach than that consider by Julián go as follows: Note that 
\begin{eqnarray}
 |x|^px-|y|^py &=& \int_0^1\frac{d}{dt}|y+t(x-y)|^{p}(y+t(x-y))dt      \nonumber \\
   &=& (x-y)\int_0^1|y+t(x-y)|^p+ \\ &&p\int_0^1 |y+t(x-y)|^{p-2}\langle y+t(x-y),x-y\rangle    (y+t(x-y))dt \nonumber 
\end{eqnarray}
Also, by using the hint, we have that \begin{eqnarray}
 |x-y|\int_0^1|y+t(x-y)|^p &\leq& |x-y|\int_0^1 C(t^p|x|^p+(1-t)^p|y|^p)   \nonumber \\
   &\leq& C|x-y|(|x|^p+|y|^p) \nonumber 
\end{eqnarray}
On the other hand 
\begin{eqnarray}
 \int_0^1 |y+t(x-y)|^{p-2}\langle y+t(x-y),x-y\rangle    (y+t(x-y))dt &\leq&  \int_0^1 |y+t(x-y)|^p|x-y|  \nonumber 
\end{eqnarray}
and you can use the same reasoning as above. From here you can conclude
