# an Integral Inequality about harmonic function

Let $$u$$ be a real-valued harmonic function on $$\mathbb{D}$$, which extends continuously to the boundary. I wonder how to prove the inequality $$\int_\mathbb{D} e^{2u} dxdy\leq\dfrac{1}{4\pi}\left(\int_{\partial\mathbb{D}}e^uds\right)^2.$$

There are a lot of properties with harmonic function $$u$$ (like maximum principle and mean value equality), but I did not know how to handle with this $$e^u$$.

This is the isoperimetric inequality in disguise. Define the Riemannian metric with length element $$e^{u(z)}|dz|$$. The curvature of this metric is $$-e^{-2u}\Delta u=0,$$ so the metric is flat. Then your inequality says that the area of the disk is at most $$1/4\pi$$ times the square of the length of the boundary, the usual isoperimetric inequality for the flat metric.
Remark. It is sufficient to assume that curvature is non-positive, that is $$\Delta u\geq 0$$.