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Let $P(z)=(z-a)(z-b)$ where $a,b$ are any complex numbers such that $|a|\geq 1, |b|\geq 1.$ Then may I know, if $$\max_{|z|=1}|P'(z)|\leq \left(\frac{1}{2}+\frac{1}{1+|ab|}\right)\max_{|z|=1}|P(z)|$$ is true?

user159888
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    Where does that come from, and why do you think it might be true? – Martin R Aug 15 '18 at 07:50
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    That looks like useful context to be added to the question itself. – Martin R Aug 15 '18 at 08:17
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    If you are looking to make a better inequality try consider a simple test case. For example it's easy to find the maximums if we assume that $(a+b)^2/ab$ is real (arguments of the terms align if we take the argument of $x$ to be that of $-(a+b)$). Then it's not hard to see that $\frac{P'{\rm max}}{P{\rm max}} = \frac{2+|a+b|}{1+|a+b|+|ab|}$. If $|ab|=1$ then this is unity. From this it's not that hard to figure out that what you ask for is not true. – Winther Nov 04 '18 at 01:07
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    @Winther I agree with everything in your comment except the last sentence. What you have shown is that $|ab|=1$ is a limit case where the purported inequality becomes an equality. But I do not see how this prevents the purported inequality to be true outside of the limit case. – Ewan Delanoy Nov 04 '18 at 09:14

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It looks like this inequality is not fulfilled for $a=2,b=-1$:

$$\max(P'(e^{ix}))=\max{\left(5-4\cos{x}\right)^{1/2}}=3$$ $$\max(P(e^{ix}))=\max{\left(10+2\cos{x}-8\cos^2{x}\right)^{1/2}}=\frac{9}{2\sqrt{2}}$$ $$3>\left(\frac{1}{2}+\frac{1}{1+2}\right)\frac{9}{2\sqrt{2}}\approx 2.65$$

atarasenko
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