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Let $\mathbb D$ be the open unit disc $\{z:|z|<1\}\subset\mathbb C$, and suppose that $f:\mathbb D\to\mathbb D$ is holomorphic. Prove that $$\frac{|f(0)|-|z|}{1+|f(0)||z|}\le |f(z)|\le\frac{|f(0)|+|z|}{1-|f(0)||z|}.$$

My attempt: Let's first take care of the second inequality which is equivalent to prove that $$ |f(z)(1-|f(0)||z|)|\le |f(0)|+|z| .$$ We know that $$ |f(z)(1-|f(0)||z|)|\le|f(z)|+|z|, $$ but $|f(z)|$ needn't have to be less than $ |f(0)| $. How to move on? I am very confused how to deal with $f(0)$ and how to use the fact that $f$ is holomorphic over $\mathbb D$. I have also tried the Cauchy's integral formula but nothing helps. Any hint would be appreciated.


marked as duplicate by Martin R, user549397, Community Apr 23 at 8:40

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