How to show that $p-$Laplacian operator is monotone? Define 
$$\langle -\Delta_p u, v \rangle_{(W^{1,p})', W^{1,p}} = \int_{\Omega}|\nabla u |^{p-2}\nabla u \nabla v.$$
How do I show that this operator is monotone? I get
$$\langle-\Delta_p u_1 - \Delta_p u_2, u_1-u_2\rangle = \int(\nabla u_1 - \nabla u_2)(|\nabla u_1|^{p-2}\nabla u_1 - |\nabla u_2|^{p-2}\nabla u_2) $$
Now adding and subtracting $|\nabla u_1|^{p-2}\nabla u_2$ in the brackets still doesn't help with one of the terms..
(Recall an operator is monotone if $\langle Tx - Ty, x-y\rangle \geq 0.$)
 A: Let $x,y\in \mathbb{R}^N$ and note that $$\tag{1}|x|^{p-2}x-|y|^{p-2}y=\int_0^1\frac{d}{dt}\left(|y+t(x-y)|^{p-2}(y+t(x-y))\right)dt$$
By calculating the derivative, we conclude from $(1)$ that $$\tag{2}|x|^{p-2}x-|y|^{p-2}y=(y-x)\int_0^1|y+t(y-x)|^{p-2}dt+\\+ (p-2)\int_0^1|y+t(y-x)^{p-4}|\langle y+t(y-x),x-y\rangle(y+t(y-x))dt$$
It follows from $(2)$ that $$\tag{3}\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle =|x-y|^2\int_0^1|y+t(y-x)|^{p-2}dt+\\ +(p-2)\int_0^1|y+t(y-x)|^{p-4}\langle y+t(y-x),x-y\rangle^2dt$$
If $p\geq 2$, then we conclude the positivity from $(3)$. If $p<2$ we note that 
$$\int_0^1|y+t(y-x)|^{p-4}\langle y+t(y-x),x-y\rangle^2dt\leq|x-y|^2\int_0^1|y+y(y-x)|^{p-2}$$
which again implies the positivity. I would like to remark that in fact we have a more Strong inequality (which is used, for example, in showing that $-\Delta_p$ is $S_+$)
$$\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle \geq \left\{ \begin{array}{cc}
c_p|x-y|^p  &\mbox{ if $p\geq 2$} \\
 c_p\frac{|x-y|^2}{(|x|+|y|)^{2-p}}  &\mbox{if $p<2$}
       \end{array} \right.
$$
where $c_p>0$ is a constant. For this strongly inequality see the notes of Ireneo Peral in the Appendix.
A: We have, following Keeran's idea from the comments - for $p \ge 2$: 
$\def\sp#1{\left\langle#1\right\rangle}\def\abs#1{\left|#1\right|}\def\Lp{\Delta_p}\def\np#1{\abs{\nabla #1}^{p-2}}$
\begin{align*}
   \sp{-\Lp u_1 + \Lp u_2, u_1 - u_2} &= \int_\Omega \bigl(\np{u_1}\nabla u_1 - \np{u_2}\nabla u_2\bigr)(\nabla u_1 - \nabla u_2)\\
   &= \int_\Omega \abs{\nabla u_1}^p - \bigl(\np{u_1}+\np{u_2}\bigr)\nabla u_1\nabla u_2 + \abs{\nabla u_2}^p\\
   &\ge \int_\Omega \abs{\nabla u_1}^p + \abs{\nabla u_2}^p - \frac 12\bigl(\np{u_1} + \np{u_2}\bigr)\bigl(\abs{\nabla u_1}^2 + \abs{\nabla u_2}^2\bigr)\\ \def\nn#1#2{\abs{\nabla #1}^{#2}}
   &= \frac 12 \int_\Omega  \nn{u_1}p + \nn{u_2}p - \np{u_1}\nn{u_2}2 -  \np{u_2}\nn{u_1}2\\
   &= \frac 12 \int_\Omega \bigl(\np{u_1} - \np{u_2}\bigr)\bigl(\nn{u_1}2 - \nn{u_2}2\bigr)\\
   &\ge 0.
\end{align*}
For the final inequality we used that $x\mapsto x^2$ and $x \mapsto x^{p-2}$ are monotonically increasing on $\mathbb R$.
A: Although the answers  given by other ones are good enough, I would like to give a very simple proof the inequality below: 
$$
\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle\geq 0, \quad 1<p<\infty.
$$
which can induce the desired result and only takes advantage of Young's inequality. 
By a direct computation, 
$$
\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle=|x|^{p}+|y|^p-|x|^{p-2}x\cdot y
-|y|^{p-2}y\cdot x,
$$
By Young's inequality, 
$$
||x|^{p-2}x\cdot y|\leq |x|^{p-1}|y|\leq \frac{|x|^p}{p'}+\frac{|y|^p}{p},
$$
where $p'=\frac{p}{p-1}$. Similarly, $||y|^{p-2}y\cdot x|\leq \frac{|y|^p}{p'}+\frac{|x|^p}{p}$. Hence, 
$$
-|x|^{p-2}x\cdot y
-|y|^{p-2}y\cdot x\geq -|x|^p-|y|^p.
$$
Hence, $$
\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle\geq 0. 
$$
Of course, this inequality fails to be optimal but good enough to show that p-Laplacian is monotone. 
