If $P(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n$ is a polynomial of degree $n\geq 1$ having all its zeros in $|z|\leq 1,$ then for all $z$ on $|z|=1$ for which $P(z)\neq 0$ $$\text{Re}\left(\frac{zP'(z)}{P(z)}\right)\geq \frac{n-1}{2}+\frac{1}{1+|a_0|}$$ is true. Refer An inequality involving the logarithmic derivative of a polynomial
Can we improve this with the same hypothesis to $$\text{Re}\left(\frac{zP'(z)}{P(z)}\right)\geq \frac{n-2}{2}+\frac{1}{1+|a_0|}+\frac{n}{n+1}?$$
No you can't. Let $P(z) = z^n-1$ for $n \ge 1$ odd. Then, for $z = -1$, $n\frac{1}{2} = Re(z\frac{nz^{n-1}}{z^n-1}) \ge^? \frac{n-2}{2}+\frac{1}{2}+\frac{n}{n+1}$. This holds iff $\frac{n}{n+1} \le \frac{1}{2}$, which is false for large $n$.