Derivative of a polynomial $P(z)$ and the derivative of the conjugate reciprocal of $P(z)$

Question

Let $P(z)=\sum_{n=0}^na_nz^n$ be a polynomial of degree $n$ having no zeros in $|z|<1.$ Let $Q(z)=z^n\overline{P(1/\overline{z})}.$ Then it is an easy exercise to show that $\Re\left(zP'(z)/P(z)\right)= \sum_{k=1}^n\Re\frac{z}{z-z_k}\leq n/2$ on $|z|=1,$ since each $\Re\frac{z}{z-z_k}\leq 1/2$ on $|z|=1.$ But using the property $\Re (z)\leq 1/2$ iff $|z|\leq |1-z|$ with little simplification we will have then on $|z|=1,$ $$|P'(z)|\leq |nP(z)-zP'(z)|=|Q'(z)|.$$ Suppose we keep reducing the radius of the disc containing no zeros inside, what may happen to this inequality? Is there any closed form relation between $|P'(z)|$ and $|Q'(z)|$ on $|z|=1?$ For example $P(z)=z^n+K^n, K\leq 1,$ we have the equality relation $K^n|P'(z)|= |Q'(z)|$ on $|z|=1.$ This makes us feel that for a polynomial $P(z)$ having no zeros in $|z|<K, K\leq 1$there may be a closed expression of the kind $f(K)|P'(z)|\leq |Q'(z)|$ on $|z|=1,$ where $f(K)$ is positive and $f(K)\rightarrow 0$ as $K\rightarrow 0.$ What may be that $f(K)?$ If the polynomial has all its zeros on $|z|=K,$ then $|P'(z)|K^n\leq |Q'(z)|.$ I think it should be of something like $f(K)=\frac{K^n}{1+((|a_0|/|a_n|)-K^n)(1-K^n)}.$ It is purely by intution, I may be incorrect also!