Subharmonic functions Consider:

if $\log u$ and $\log v$ are both subharmonic function on an open set $U$, then $\log(u+v)$ is also subharmonic on $U$.

How can I prove this?
 A: If $w$ is strictly positive and $C^2$ then a direct calculation shows that $\log w$ is subharmonic if and only if $w\Delta w+|\nabla w|^2 \ge 0$.  Taking $w=u+v$ we see that $\log(u+v)$ is subharmonic if and only if
$$
(u\Delta u+|\nabla u|^2)+(v\Delta v+|\nabla v|^2) +\Delta(uv) \ge 0.
$$
If $\log u$ and $\log v$ are subharmonic then  the first two terms on the left are non-negative. And so is the third because 
$$
\Delta(uv) = uv\cdot\left[\Delta(\log u+\log v) +|\nabla(\log uv|^2\right],
$$
which is clearly non-negative.
A: The assumption about the regularity is not necessary here.
Indeed, we know that we can find a sequence of smooth functions $u_n$ and $v_n$ decreasing to $u$ and $v$ respectively and such that $\log u_n, \log v_n$ are subharmonic (just use a convolution with a radial smoothing kernel).
We may then use the argument of JohnD which guarantees that $\log(u_n+v_n)$ is subharmonic. But a limit of a decreasing sequence of subharmonic functions is still subharmonic, so that $\log(u+v)$ is subharmonic.
Moreover, a direct proof (with no assumption on regularity) may be obtained once we know some basic stuff about sh functions. Namely, $f(x,y)=\log (e^x+e^y)$ defines a a convex function, non-decreasing in each variable, therefore $f(\log u, \log v)=\log(u+v)$ is subharmonic. 
