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Let $f(x)\in\mathbb{R}[x]$ be a polynomial of degree $n$, which has only real zeros. I would like to show that $$(n − 1)(f'(x))^2 \geq nf(x)f''(x),$$ where $f'$ and $f''$ denote the first and second derivative of $f$, respectively.

What I have tried so far:

1) Let $z_1,\ldots, z_n$ be the zeros of $f$, then the inequality follows for these zeros of $f$.

2) Equality is possible for these zeros of $f$ only if $f'(z_i)=0$ too.

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    $\begingroup$ @coffeemath yes, thanks. Fixed now. Regarding your second question, it is not stated in the problem. In what I have tried, I just assumed that those are real roots of $f$, there might be $i\neq j$ such that $z_i=z_j$. $\endgroup$ – Luckyluck63 May 3 '16 at 8:11
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Here's an outline of how I got to this, rather than a pretty, cleaned up proof.

First I worked out that $$ (-\log{f})'' = \frac{f'^2}{f^2}-\frac{f''}{f}, $$ which I thought might be useful because if you divide by $f^2 \geqslant 0$, you find terms that look like $f'/f$. Therefore the inequality is equivalent to showing $$ n (-\log{f})'' \geqslant (-\log{f})'^2. $$ What are each of these for a polynomial? Well, if $f(x)=K\prod_{i=1}^n (x-a_i)$, $$ -\log{f(x)} = -\log{K}-\sum_{i=1}^n \log{(x-a_i)}, $$ whether or not the $a_i$ are distinct. Then $$ (-\log{f(x)})' = \sum_{i=1}^n \frac{-1}{x-a_i}, $$ and so $$ (-\log{f(x)})'^2 = \left( \sum_{i=1}^n \frac{-1}{x-a_i} \right)^2 \leqslant \sum_{i=1}^n \frac{1}{(x-a_i)^2} \cdot \sum_{i=1}^n 1 = n(-\log{f(x)})'', $$ by Cauchy-Schwarz, and this is what we wanted.

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