Show $|f(1/3)|\ge 1/5$

Question

Let $f$ be analytic on the unit disc $D$ and $|f(z)|\le 1$ with $f(0)=1/2$. Show $|f(1/3)|\ge 1/5$.

I use Schwarz Lemma

$$\left |\frac{f(z)-f(0)}{1-\overline{f(0)}f(z)}\right |\le \left | \frac{z-0}{1-\bar{0}z} \right |$$

And so I get $3|2f(1/3)-1| \le |2-f(1/3)|$ for $z=1/3$ but I’m having trouble extracting the right bound from this. What’s the easiest way?

Answer 1

I think there is some mistake, anyway, it looks like this: \begin{align*} \dfrac{1}{3}&\geq\dfrac{|f(1/3)-(1/2)|}{|1-(1/2)f(1/3)|}\\ 3|f(1/3)-(1/2)|&\leq|1-(1/2)f(1/3)|\\ 3((1/2)-|f(1/3)|)&\leq 1+(1/2)|f(1/3)|\\ 1/2&\leq(7/2)|f(1/3)|\\ |f(1/3)|&\geq1/7. \end{align*}