Let $u\in C^{\infty}(\mathbb{R}^3)$. Prove that seeing as distributions,
$$\triangle |u|\ge \triangle u \cdot \text{sign}(u)$$
which means, $\forall \phi$ smooth and compactly supported on $\mathbb{R}^3$, $\phi \ge 0$, we have
$$\int \triangle |u|(x) \phi(x) \ge \int \triangle u(x) \cdot \text{sign}(u(x))\phi (x)$$
where $\text{sign}(u)$ is the sign function of $u$, equals 1 when $u(x)>0$, -1 when $u(x)<0$, 0 when $u(x)=0$.
I have tried to deal with Laplace operator and change the inequality into form $\int \text{sign}(u) (u\triangle \phi-\phi\triangle u)\ge 0$, but still don't know where to go on. Thanks for any help.
Set $V=\{u < 0\}$ and $W = \{u > 0\}$. Then
$$\int \Delta u \cdot \text{sign}(u) \phi\, dx = \int_{W} \Delta u \phi \, dx - \int_V \Delta u \phi \, dx.$$
Apply Green's identities to get
$$\int \Delta u \cdot \text{sign}(u) \phi\, dx = \int |u| \Delta \phi \, dx + \int_{\partial W} \frac{\partial u}{\partial \nu_1} \phi \, dx -\int_{\partial V} \frac{\partial u}{\partial \nu_2} \phi \, dx$$
where $\nu_1$ is the outward normal to $W$ and $\nu_2$ is the outward normal to $V$. Since $\partial u/\partial \nu_1 \leq 0$, $\partial u/\partial \nu_2 \geq 0$, and $\phi \geq 0$ we have
$$\int \Delta u \cdot \text{sign}(u) \phi\, dx \leq \int |u| \Delta \phi \, dx.$$