TITLE: How to show that $p-$Laplacian operator is monotone?
QUESTION [7 upvotes]: 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 $\langle Tx - Ty, x-y\rangle \geq 0.$)

REPLY [5 votes]: 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.