Let $P(z)=\sum_{m=0}^na_mz^m$ be a polynomial of degree $n\geq 1$ having no zeros in $|z|<1,$ then for any complex number $\alpha$ with $|\alpha|=1,$ is it true on $|z|=1$ that $$\left|\alpha zP'(z)+zP'(z)-\left(n-\frac{|a_0|-1}{|a_0|+1}\right)P(z)\right|\leq \left(n-\frac{|a_0|-1}{|a_0|+1}\right)\max_{|z|=1}|P(z)|?$$ The inequality arises from the case of $\alpha=-1.$
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$\begingroup$ This is wrong. The RHS is negative for $P(z)=2$. $\endgroup$– Philipp LampeCommented Sep 4, 2018 at 12:33
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1$\begingroup$ Let $Q(z) = 2 + z$. Consider $P(z) = \lambda Q(z)$ for $\lambda \in \mathbb{R}$. Let $\alpha = 1$. Evaluating the left hand side on $z = 1$ gives $| 2\lambda - 6 \lambda /(2\lambda + 1)|$. The right hand side is exactly $6\lambda / (2\lambda + 1)$. As $\lambda \nearrow +\infty$ you see that your inequality is false. You probably want $n \geq 2$ at least. $\endgroup$– Willie WongCommented Sep 6, 2018 at 4:29
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