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Assume $f:D\to D$ is holomorphic and $f(a) = f(b) = 0$ for $a\neq b$ and $a,b\in D$, where $D = D(0,1)$. Prove that $$|f(z)| \leq \left| \frac{z-a}{1-\bar{a}z}\right| \left|\frac{z-b}{1-\bar{b}z}\right| $$

In order to use Schwarz lemma, I take $\varphi_{-a}\in\text{Aut}(D)$ where $$\varphi_{-a}(z) := \frac{z+a}{1+\bar{a}z}$$ then $\varphi_{-a}(0) = a$ and it's easy to prove that $\varphi_{-a}\circ\varphi_a = \text{id}_D$.

Now let $g = f\circ\varphi_{-a}$ which is holomorphic on $D$, then $g(0) = f(\varphi_{-a}(0)) = f(a) = 0$, and by Schwarz lemma we have $|g(w)|\leq |w|$. Take $w = \varphi_a$ then we get $|g(\varphi_a(z))|\leq |\varphi_a(z)|$, i.e. $$|f(\varphi_{-a}(\varphi_a(z)))| = |f(z)|\leq |\varphi_a(z)|=\left|\frac{z-a}{1-\bar{a}z}\right|$$ and similarly $$|f(z)|\leq \left|\frac{z-b}{1-\bar{b}z}\right|$$ Hence $$|f(z)|^2\leq \left| \frac{z-a}{1-\bar{a}z}\right| \left|\frac{z-b}{1-\bar{b}z}\right|$$ But $|f(z)|<1$ which means that $|f(z)|^2<|f(z)|$, so this proof seems to fail...

How can I solve it then?

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You have gotten $$ |f(z)|\leq \left|\frac{z-a}{1-\bar{a}z}\right|.\tag{1}$$ Consider $$F(z)=\left(\frac{1-\bar{a}z}{z-a}\right)f(z),$$ which satisfies $|F(z)|\le 1$ by $(1)$. Therefore $z=a$ is a removal singularity of $F(z)$. So $F:D\to D $ is holomorphic and $F(b)=0.$ You can get $$|F(z)|\leq \left|\frac{z-b}{1-\bar{b}z}\right|,$$ which yields $$ |f(z)| \leq \left| \frac{z-a}{1-\bar{a}z}\right| \left|\frac{z-b}{1-\bar{b}z}\right|.$$

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