For any polynomial $P_{n}(x)$ with degree $n$ that all its zeros lie in$[-1,1]$, then $\|P_{n}'(x)\|>C \sqrt{n} \|P_{n}(x)\|$ where $C$ is an absolute constant.
2026-03-29 17:31:12.1774805472
How to prove Turan's Inequality
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What you have seems to be incorrect. For instance, consider $$P_2(x) = (x-a)(x-b)$$ Then $P_2'(x) = 2x - (a+b)$. Consider $x^* = \dfrac{a+b}2$. Then we have $$P_2'(x^*) = 0$$ whereas $$P_2(x^*) = \left(\dfrac{b-a}2\right)\left(\dfrac{a-b}2\right) = - \dfrac{(a-b)^2}4$$ which clearly violates the inequality.