I'm trying to show that for a weak solution of the p-laplacian dirichlet problem, that is, for $u \in W^{1.p}_0 (\Omega)$ s.t. $$ \int_\Omega |\nabla u|^{p-2} \nabla u \cdot \nabla v = \int_\Omega f v \ \text{, } \forall v \in W^{1.p}_0 (\Omega) $$ then $u$ minimizes the energy functional associated $$ \Phi(u) = \int_\Omega \frac{|\nabla u|^p}{p} - fu $$
The other direction, that is, showing that $\Phi^\prime (u)(v) = 0$, was ok to do, so i know that $u$ is a critical point, but i don't know to show that is indeed a minimum point.
Any hints are appreciated.
Ok so the minimization is easily attainable once one has the following inequality
$$ |b|^p \geq |a|^p + p<|a|^{p-2}a,b-a> $$
as shown in here.
Then, taking $b=\nabla v$ and $a = \nabla u$, and therefore
$$ \frac{1}{p}|\nabla v|^p \geq \frac{1}{p}|\nabla u|^p + <|\nabla u|^{p-2}\nabla u,\nabla (v-u)> $$
so integrating both sides yields
$$ \frac{1}{p} \int |\nabla v|^p \geq \frac{1}{p} \int|\nabla u|^p + \int |\nabla u|^{p-2}\nabla u \cdot\nabla (v-u) $$ as $v-u \in W_0^{1,p}$ and $u$ is a weak solution
$$ \int |\nabla u|^{p-2}\nabla u \cdot\nabla (v-u) = \int f(v-u) $$
then it follows that
$$ \frac{1}{p} \int |\nabla v|^p - \int fv \geq \frac{1}{p} \int|\nabla u|^p -\int fu $$
as desired.