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Let $|z|=r$, apply the Cauchy estimate to the disc $|\zeta-z|<(1-r)/2$. We obtain $$|f'(z)|\leq \frac{2}{1-r}\frac{(1+r)^2}{(1-r)(3+r)}.$$ Maximizing the factor $(1+r)^2/(3+r)$ by Calculus, we obtain that is it at most $1$.

This gives $$|f'(z)|\leq\frac{2}{(1-|z|)^2}$$ which is worse than conjectured only by a factor of $(1+|z|)^2$, which is at most $4$.

One Perhaps one can further improve the constant by applying Cauchy to a disc of radius $t\in(0,1-r)$, and then optimizing in $t$, which leads to solving a cubic equation.

Is it really important whether the constant is 4 or 2.87 ?

It is not likely that a simple extremal function exists, and I suppose that probably for each $z$ there will be a different extremal function.

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Let $|z|=r$, apply the Cauchy estimate to the disc $|\zeta-z|<(1-r)/2$. We obtain $$|f'(z)|\leq \frac{2}{1-r}\frac{(1+r)^2}{(1-r)(3+r)}.$$ Maximizing the factor $(1+r)^2/(3+r)$ by Calculus, we obtain that is it at most $1$.

This gives $$|f'(z)|\leq\frac{2}{(1-|z|)^2}$$ which is worse than conjectured only by a factor of $(1+|z|)^2$, which is at most $4$.

One can further improve the constant by applying Cauchy to a disc of radius $t\in(0,1-r)$, and then optimizing in $t$, which leads to solving a cubic equation.

Is it really important whether the constant is 4 or 2.87 ? It is not likely that a simple extremal function exists, and I suppose that for each $z$ there will be a different extremal function.