$\frac{|f(0)|-|z|}{1+|f(0)||z|}\le |f(z)|\le\frac{|f(0)|+|z|}{1-|f(0)||z|}$ in the unit disc for $f$ holomorphic. [duplicate]

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.