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 $e^u$.



an Integral Inequality about harmonic function
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 $\Delta u\geq 0$.


Ref. MR0936419 Burago, Yu. D.; Zalgaller, V. A. Geometric inequalities. Translated from the Russian by A. B. Sosinskiĭ. Springer-Verlag, Berlin, 1988.


Isoperimetric inequality for arbitrary surfaces is usually called Fiala-Aleksandrov inequality. The original publications are:


MR0006422 Fiala, F. Le problème des isopérimètres sur les surfaces ouvertes à courbure positive, Comment. Math. Helv. 13 (1941), 293–346, and


MR0052136 Aleksandrov, A. D. An isoperimetric problem, Doklady Akad. Nauk SSSR (N.S.) 50, (1945). 31–34.


