Mathematical Olympiad question (complex variable) I found this question in an old Mathematical Olympiad:

Let $0<a<1$ be a real number, and let $f(z)$ be a complex polynomial such that $$|f(z)|\leq \frac{1}{|z-a|}$$ on the unit disk $|z|\leq 1$. Prove that $$|f(a)| \leq \frac{1}{1-a^2}.$$

My attempt: Since $f$ is analytic in $\{z\in\mathbb{C}:|z|\leq 1\}$, we have that $$|f(a)| \leq \max_{|z|=1}|f(z)|.$$ Because of the triangle inequality, for all $z\in \mathbb{C}$ with $|z|=1$ we have $$|z-a|\geq |z|-a = 1-a.$$ By applying the hypothesis, we get $$|f(a)| \leq \max_{|z|=1}|f(z)| \leq \max_{|z|=1}\frac{1}{|z-a|}\leq \frac{1}{1-a}.$$ Nevertheless, we know that $1-a^2>1-a$ since $a\in (0,1)$. Therefore, we cannot get the desired result from the above inequality.
What can I apply to complete the proof?
 A: Let $F(z)=(z-a)f(z)$ and $G(z)=F(\frac {z+a} {1+az})$. [Recall that $ z \to \frac {z+a} {1+az}$ is an analytic bijection of the unit disk into itself whose inverse is $ z \to \frac {z-a} {1-az}$]. Now $|G(z)| \leq 1$ and $G(0)=F(a)=0$. By Schwarz Lemma we get $|G(z)| \leq |z|$.  This can be written as $|F(z)|\leq |\frac {z-a}{1-az}|$. Hence, $|f(z)|\leq |\frac 1{1-az}|$ and putting $z=a$ finishes the proof.
A: From Cauchy’s integral formula: $$f(a) = \frac1{2\pi \mathrm i} \int_{\lvert z\rvert =1} \frac{f(z)}{z-a}\mathrm dz.$$ Then, using that $\overline z = z^{-1}$ if $\lvert z \rvert = 1$: $$
\begin{eqnarray}
\lvert f(a) \rvert &\leq& \frac1{2 \pi} \int_{\lvert z \rvert = 1} \frac {\lvert f(z) \rvert}{\lvert z - a \rvert} \lvert \mathrm dz \rvert\\
&\leq& \frac1{2 \pi} \int_{\lvert z \rvert = 1} \frac1{\lvert z - a \rvert^2} \lvert \mathrm dz \rvert\\
 &=& \frac1{2 \pi \mathrm i} \int_{\lvert z \rvert = 1} \frac1{\lvert z - a \rvert^2} z^{-1} \mathrm dz \\
&=& \frac1{2 \pi \mathrm i} \int_{\lvert z \rvert = 1} \frac{\mathrm dz}{(1 - a z)(z - a)}\\
&=&\frac1{1-a^2}
\end{eqnarray}$$
The last equality is again Cauchy’s formula for the holomorphic function $$z \mapsto \frac1{1-a z}.$$
A: (same kind of idea as in the answer of Kavi Rama Murthy) :
When $f(z)/g(z)$ has only removable singularities in ${\Bbb D}$ and $|f(z)|\leq |g(z)|$ on $\partial{\Bbb D}$ then $|f(z)|\leq |g(z)|$ in ${\Bbb D}$.
For $|a|<1$,  we have $\left| \frac{z-a}{1-\bar{a}z}\right|=1 $ when $|z|=1$ so using the above principle and the hypotheses, $|f(z)(z-a)|\leq \left| \frac{z-a}{1-\bar{a}z}\right|$ on the unit disk. Simplifying the common factor and setting $z=a$ yields the result.
A: Complement to @WimC's nice answer
(Perhaps an explicit use of the condition: $f(z)$ is a complex polynomial.)
Let $t\in \mathbb{R}$. We have
$$\left|\frac{f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}\right|
\le \frac{1}{|\mathrm{e}^{\mathrm{i} t} - a|^2} = \frac{1}{1 + a^2 - 2a \cos t}$$
and
$$\int_0^{2\pi} \left|\frac{f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}\right|\mathrm{d} t
\le \int_0^{2\pi} \frac{1}{1 + a^2 - 2a \cos t} \mathrm{d} t
= \int_0^{\pi} \frac{2}{1 + a^2 - 2a \cos t} \mathrm{d} t = \frac{2\pi}{1 - a^2}. \tag{1}$$
(Note: Using the substitution $t = 2\arctan x$, it is easy to evaluate the integral.)
Since $f(z)$ is a complex polynomial, let $f(z) = \sum_{m=0}^n b_m z^m$.
We have
$$\frac{f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}
= \sum_{m=0}^n \frac{b_m \mathrm{e}^{\mathrm{i} mt}\mathrm{e}^{-\mathrm{i} t}}{1 - a\mathrm{e}^{-\mathrm{i} t}}
= \sum_{m=0}^n b_m \mathrm{e}^{\mathrm{i} mt}\mathrm{e}^{-\mathrm{i} t}
\sum_{k=0}^\infty (a \mathrm{e}^{-\mathrm{i} t})^k$$
and
$$\frac{\mathrm{e}^{\mathrm{i} t}f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}
= \sum_{m=0}^n b_m
\sum_{k=0}^\infty a^k \mathrm{e}^{\mathrm{i}(m - k) t}$$
and
$$\int_0^{2\pi} \frac{\mathrm{e}^{\mathrm{i} t}f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}\mathrm{d} t
= \sum_{m=0}^n b_m 
\sum_{k=0}^\infty a^k \int_0^{2\pi} \mathrm{e}^{\mathrm{i}(m - k) t}\mathrm{d} t = \sum_{m=0}^n 2\pi b_m a^m = 2\pi f(a).$$
Using (1), we have
$$|f(a)| = \frac{1}{2\pi} \left|\int_0^{2\pi} \frac{\mathrm{e}^{\mathrm{i} t}f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}\mathrm{d} t\right| \le \frac{1}{2\pi} \int_0^{2\pi} \left|\frac{\mathrm{e}^{\mathrm{i} t}f(\mathrm{e}^{\mathrm{i} t})}{\mathrm{e}^{\mathrm{i} t} - a}\right|\mathrm{d} t \le \frac{1}{1 - a^2} .$$
We are done.
