One way to derive the global inequality is the Principle of Harmonic majorant: If $D$ is a bounded region, $u$ is harmonic in $\overline{D}$, $f$ is analytic in $\overline{D}$, then the inequality $\log|f(z)|\leq u(z),\, z\in\partial D$ implies the same in $D$. To prove this, just apply the Maximum principle to the harmonic function $\log|f|-u$ in the region $D$ minus small neighborhoods of zeros of $f$, that the inequality holds on the boundaries of these small neighborhoods is clear since $\log|z|\to-\infty$ when $z\to$ a zero of $f$.

Now to prove your global inequality, use as $u$ the solution of the Dirichlet problem with the boundary values $\log|f|$.

The solution of the Dirichlet problem for a disk is completely elementary, and the value at the center is just the average of the values on the circumference.

So we only used the Maximum Principle for harmonic functions and solvability of the Dirichlet problem for a disk.

One way to derive the global inequality is the Principle of Harmonic Majorant: If $D$ is a bounded region, $u$ is harmonic in $\overline{D}$, $f$ is analytic in $\overline{D}$, then the inequality $\log|f(z)|\leq u(z),\, z\in\partial D$ implies the same in $D$. To prove this, just apply the Maximum principle to the harmonic function $\log|f|-u$ in the region $D$ minus small neighborhoods of zeros of $f$; that the inequality holds on the boundaries of these small neighborhoods is clear since $\log|z|\to-\infty$ when $z\to$ a zero of $f$.

Now to prove your global inequality, use as $u$ the solution of the Dirichlet problem with the boundary values $\log|f|$.

The solution of the Dirichlet problem for a disk is completely elementary, and the value at the center is just the average of the values on the circumference.

So we only used the Maximum Principle for harmonic functions and solvability of the Dirichlet problem for a disk.