Let $f$ be a function analytic at the set $\overline{Ball_r(0)}$,and assume $f$ is bounded on $\left\{|z|=r\right\}$ so that $|f|\le a$ for the upper half circle, and $|f|\le b$ for the lower half circle. I wish to prove that $|f(0)|\le \sqrt{ab}$
The problem can be solved using Cauchy integral, but I solved it using the maximum principle, and would like to know if I done that right.
Here is what I did:
Let $g(z)=f(z)f(-z)$, this function is also analytic on $\overline{Ball_r(0)}$ as a composition of analytic functions. If $z$ is on the upper or lower circle we get that $|f(z)f(-z)|\le |f(z)||f(-z)|\le ab$ so for $z=0$, we get $|f(0)^2|\le ab \rightarrow |f(0)|\le \sqrt{ab}$
but please let me know if this solution is good.
Your solution is fine, and optimal in this situation.
An alternative is to use Jensen's formula (or the fact that $\log|f(z)|$ is subharmonic): $$ \log|f(0)| \le \frac{1}{2\pi} \int_0^{2\pi} \log|f(re^{it})| \, dt \le \frac{\pi \log a + \pi \log b}{2\pi } = \frac{\log a + \log b}{2}. $$
This works also if there are $n$ different bounds on $n$ arbitrary arcs on the circle (whose union is the entire circle).