I have been thinking about this problem off and on for a couple days now to no avail.
If in answering this question you could address the things I've tried so far as "on the right track", "not really a good idea", etc, I would be very grateful.
Let $u: \mathbb{C} \to \mathbb{R}$ be a harmonic function on the entire complex plane, and let $u_+$ be the positive part, that is, $u_+(z) = \max(u(z), 0)$. Let $$T(r) = \int_{0}^{2\pi} u_+(re^{i\theta}) d\theta$$ Show that $T(r) \to \infty$ as $r\to \infty$.
It's so bothersome to me because it seems like an elementary and intuitive result.
My first thought, purely based on mental pictures I have of harmonic functions, was that a harmonic function with $u(0) = 0$ (the problem can be reduced to this) has some sort of radial monotonicity or even convexity, ie, that $r\mapsto u(re^{i\theta})$ is convex for $r \in [0,1]$. This would surely give the result; the only problem is that it's very false.
My other idea was as follows: if $R> r$, let $A = \{\theta \in [0, 2\pi): u(re^{i\theta}) > 0\}$. If we can get some constant $C_{R/r}>1$ which satisfies $$\int_{\theta \in A} u(re^{i\theta}) d\theta \leq C_{R/r} \int_{\theta \in A} u(Re^{i\theta}) d\theta$$ then we'll be done, since $$T(r) \leq C_{R/r} \int_{\theta \in A} u(Re^{i\theta}) d\theta \leq C_{R/r}T(R)$$
The obvious thing to try to make some estimate like this work is to represent $u(re^{i\theta}) = P_{r/R} * u(Re^{i\theta})$, where $P_q$ is the Poisson kernel. But I have not been able to make this work.
I don't think this problem should be this hard, so please point me in the right direction. Thanks!