Past exam of the fixed point methods for nonlinear partial differential equations course of the University of Oxford / 2008 / 4 / b / Solution

For any $v\in M$, the segment $u+t(v-u)$ stays in $M$ for $0\le t\le 1$. Since $u$ is a minimizer, the right derivative of its energy at $t=0$ is nonnegative. The first variation therefore gives the variational inequalityThe factor two comes from the quadratic gradient term. Only a one-sided inequality follows because a variation pointing below the obstacle need not be admissible.