# A New Analytic Inequality

Consider an analytic function $f : U \longrightarrow \mathbb{C}$ where $U$ is an open subset of the complex numbers which contains the closed unit disk. I have $|f(x)| \geq 1$ for any $x \in [-1, 1]$. Is the following true ? $$|\int_0^{2\pi} f(e^{it})\overline{f(e^{-it})}dt| \geq 4$$

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Yes, indeed $$\frac{1}{2\pi}\int_0^{2\pi} f(e^{it})\overline{f(e^{-it})}dt= |f(0)|^2,$$ as you can check applying the Cauchy formula to the holomorphic function $f(z)\overline{f(\bar z)}$.