How do I prove this well known inequality? Let $u$ and $v$ be functions from $\mathbb{R}^{n+1}$ to $\mathbb{C}$.
How do I prove the following inequality:
$$\big| |u|^{p-1}u - |v|^{p-1}v\big| \leq (|u|^{p-1}+|v|^{p-1})|u-v|?$$
This inequality is often used showing well-posedness of some non-linear PDE without proof.
EDIT: Some constant which is depending on $p$ and $n$ may be multiplied on RHS, i.e.
$$\big| |u|^{p-1}u - |v|^{p-1}v\big| \leq C(n,p)(|u|^{p-1}+|v|^{p-1})|u-v|.$$
 A: Observe
\begin{align}
\left||u|^{p-1}u-|v|^{p-1}v\right| =&\ \left||u|^{p-1}u-|u|^{p-1}v+|u|^{p-1}v -|v|^{p-1}v\right| \\
\le&\ |u|^{p-1}|u-v| + \left||u|^{p-1}-|v|^{p-1}\right||v|\\
\le&\ |u|^{p-1}|u-v|+(|u|^{p-2}|v|+|u|^{p-3}|v|^2+\ldots+|u||v|^{p-2}+|v|^{p-1})\left||u|-|v|\right|\\
\le&\ |u-v|\left(|u|^{p-1}+|u|^{p-2}|v|+\ldots +|u||v|^{p-2}+|v|^{p-1} \right).
\end{align}
Finally, we see that
\begin{align}
|u|^{a+b}+|v|^{a+b}-|u|^a|v|^b-|u|^b|v|^a =(|u|^a-|v|^a)(|u|^b-|v|^b)\geq 0
\end{align}
which means
\begin{align}
\left||u|^{p-1}u-|v|^{p-1}v\right| \leq C|u-v|\left| |u|^{p-1}+|v|^{p-1}\right|
\end{align}
for some constant $C>0$.
Note that I have assumed $p$ is an integer. It's not hard to prove the statement for more general $p$.  
Edit: Here is a more complete answer. For reference, one can consult Theorem 41 on page 39 of Inequalities by Hardy, Littlewood, and Polya.
There they obtained estimates of the form
\begin{align}
\left||u|^p-|v|^p \right|\leq p(|u|^{p-1}+|v|^{p-1})|u-v|.
\end{align}
To relate to the desired estimate to the above estimate, observe that if $u$ and $v$ are of different sign then we have
\begin{align}
\left||u|^{p-1}u-|v|^{p-1}v\right| =& |u|^p+|v|^p\leq |u|^{p-1}(|u|+|v|) \\ =&\  |u|^{p-1}|u-v|\le (|u|^{p-1}+|v|^{p-1})|u-v|.
\end{align}
If $u$ and $v$ are of the same sign then this reduces to the situation considered in the book.  
A: The inequality is false, at least for $p>2$. Let $u(x)=2$ and let $v(x)=1$. Then the left side equals $2^p-1$ and the right side equals $2^{p-1}+1$, however
$$
(2^p-1)-(2^{p-1}+1)=2^{p-1}-2
$$
which is positive when $p>2$.
A: Proceeding with a multiplicative constant in light of the other 2 answers. I'll assume $u\in\mathbb R^d$, one can make an identification $\mathbb R^2 = \mathbb C$. These things like $f(a)-f(b)=g(a,b)|b-a|$ look like MVT which usually works out.
Set $f_j(u)=|u|^{p-1}u_j$, then $\partial_i f_j=(p-1)|u|^{p-2}\frac{u_j}{|u|}u_i + |u|^{p-1}\delta_{ij}$. Lazy calculations yield
$$|\nabla f|=\|\partial_i f_j\|_{\ell^2_{ij}}\le C\|\nabla f\|_{\ell^\infty_{ij}} \le C[(p-1) |u|^{p-3}\|u_i\|^2_{\ell^\infty_{i}} + |u|^{p-1}] \le Cp|u|^{p-1},$$
since $\|u_i\|_{\ell^\infty_i} \le \|u_i\|_{\ell^2_i} = |u|$. I'd guess a smarter calculation would give $C=1$, instead we have $C=C(d)$.
Now MVT gives
$$ | |u|^{p-1}u - |v|^{p-1}v |  \le \sup_{\theta \in [0,1]} |\nabla f(\theta u + (1-\theta) v)| |u-v|$$
So the goal is just to bound the derivative on the line $[u,v]\subset \mathbb R^n$. $p=1$ is trivial. If $1<p<2$,  then $s\mapsto s^{p-1}$ is concave, and $0^{p-1}=0$. This implies subadditivity, giving the result:
\begin{align} 
|\nabla f(\theta u + (1-\theta) v)| 
&= C(n,d,p)|\theta u + (1-\theta) v|^{p-1} 
\\
&\le C(n,d,p)(|\theta u| + |(1-\theta)v|)^{p-1} \\
&\le C(n,d,p)(|\theta u|^{p-1} + |(1-\theta)v|^{p-1}) \\
&\le C(n,d,p)(|u|^{p-1} + |v|^{p-1}). 
\end{align}
For $p\ge 2$, we can just directly use convexity, like  the question body here 
and with some more consideration of constants here.
