Suppose $u \in C^2(\R^n)$  is a harmonic function.  Prove that $v=|\nabla u|^2$ is subharmonic, i.e. $-\Delta v \leq 0$

First we prove the following two lemma.

Lemma 1. Let $u$ be a harmonic function. Then $u^2$ is subharmonic.
Proof. 
\[\frac{\partial}{\partial x_i}(u^2)=2u\frac{\partial u}{\partial x_i}.\]
So 
\[\frac{\partial^2}{\partial x_i^2}(u^2)=\frac{\partial}{\partial x_i}(2u\frac{\partial u}{\partial x_i})=2u\frac{\partial^2 u}{\partial x_i^2}+2(\frac{\partial u}{\partial x_i})^2.\]
Hence
\[\Delta (u^2)=\sum_{i=1}^{n}2u\frac{\partial^2 u}{\partial x_i^2}+2(\frac{\partial u}{\partial x_i})^2=2u\Delta u+2|\nabla u|^2=2|\nabla u|^2\geq 0,\]
where we have used $\Delta u =0$ to obtain the last equality. Therfore 
\[-\Delta (u^2) \leq 0.                     \Box\]

Lemma 2. If $u$ is harmonic, then $u_{x_i}$ is also harmonic. 
Proof. We have 
\[\Delta (u_{x_i})=\frac{\partial }{\partial x_i}(\Delta u)=\frac{\partial }{\partial x_i}(0)=0.               \Box\]

Now we prove that $v=|\nabla u|^2$ is subharmonic if $u$ is a harmonic function. Note that 
\[v=\sum_{i=1}^{n}(u_{x_i})^2\]
and by Lemma 1 and Lemma 2 we know that $(u_{x_i})^2$ is a subharmonic function. Thus $v$ is subharmonic. This finishes the proof.