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Given odd positive integer $n$ and a monic polynomial $f(x)=(x-x_1)\dots (x-x_n)$ with $n$ distinct real roots. Is it always true that $\sum f'(x_i) > 0$? I may prove it for $n=3$ and $n=5$ and it looks plausible.

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A possible reformulation: given an unreduced tridiagonal matrix $\mathbf{T}$ of odd order with characteristic polynomial $f(x)$, is it true that $\mathrm{trace}(f^{\prime}(\mathbf{T}))>0$? –  J. M. Oct 10 '10 at 11:06
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up vote 16 down vote accepted

If I'm not mistaken this is basically the same question as this question from the international mathematical olympiad in 1971. The statement is only true for 3 and 5 variables showing that there is no obvious generalization to Schur's inequality in many variables.

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Instead, the sum of reciprocals $$\sum\frac{1}{f'(x_j)}$$ vanishes. This is because of the formula $$\frac{1}{f(x)}=\sum_j\frac{a_j}{x-x_j},\qquad a_j:=\frac{1}{f'(x_j)},$$ together with the asymptotics as $x\rightarrow\infty$. This is valid for every degree, odd or even.

When $n=3$, this gives an amazing proof of the property that you quote. Denote $y_j=f'(x_j)$. Then $y_1y_2+y_3y_1+y_2y_3=0$, which means that $y=(y_1,y_2,y_3)$ belongs to a quadric whose intersection with the plane $y_1+y_2+y_3=0$ reduces to $(0,0,0)$, not equal to $y$. By continuity and connexity of the parameter space $x_1< x_2< x_3$, the expression $y_1+y_2+y_3$ must keep a constant sign, which we may calculate with $f(x)=x(x^2-1)$.

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