Let $u\colon \mathbb{R}^n \to \mathbb{R}$ be a smooth function. How can we prove that $$ -\Delta\left(\lvert\nabla u\rvert^2\right) \leq 2 \nabla u \cdot \nabla\left(-\Delta u\right) $$ holds true?
My question is related to the following fact: if $u$ is harmonic in $\Omega$, i.e $-\Delta u=0$ in $\Omega$, then the previous inequality implies that $\lvert \nabla u\rvert^2$ is sub-harmonic in $\Omega$.
Just do the computations. You will find $$ \Delta\bigl(|\nabla u|^2\bigr)=2\sum_{i,j=1}^n\Bigl(\frac{\partial^2u}{\partial x_i\partial x_j}\Bigr)^2+2\,\nabla u\cdot\nabla\bigl(\Delta u\bigr). $$