Let $a_i>0$, $i=1,\dots,n$, and put $\overline{a}:=\frac{1}{n}\sum_{i=1}^n a_i$. Assuming not all $a_i$'s are equal, take $$ p(x):=\sum_{i=1}^n a_i (a_i-\overline{a})\prod_{k=1,\dots,n\;k\neq i} (x+a_k)^2. $$ How to prove that $p(x)$ has exactly one positive root? (this is a conjecture, based on numerical experiments)

  • $\begingroup$ What makes you tho expect that it has exactly one positive root? $\endgroup$ – Seva May 3 '15 at 8:58
  • $\begingroup$ @Seva some numerical evidence, I'll add this to the question $\endgroup$ – dima May 3 '15 at 8:59
  • $\begingroup$ Can you prove it for $n=1$, $n=2$, $n=3$? $\endgroup$ – Greg Martin May 3 '15 at 19:21
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    $\begingroup$ If you divide by $\prod (x+a_k)^2$, you can replace the product with the reciprocal of the omitted term, $(x+a_i)^{-2}$. $\endgroup$ – Douglas Zare May 3 '15 at 20:07
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    $\begingroup$ For $n=3$ and also for $n=4$ it seems there is just one change of sign in the consecutive terms, so Descartes' rule of signs should kick in and yield at most one positive root. Perhaps one can even inductively show that there is just one change of sign, and a small additional argument shows that there is at least one positive root. $\endgroup$ – Suvrit May 4 '15 at 4:25

First, as suggested by Douglas Zare, we can rewrite $p(x)$ as $$p(x) = \prod_{i=1}^n (x+a_i)^2 \cdot \sum_{i=1}^n \frac{a_i(a_i-\overline{a})}{(x+a_i)^2}.$$ Notice that the number of positive roots of $p(x)$ equals the number of positive roots of $$q(x):=\frac{(x+\overline{a})^2}{\prod_{i=1}^n (x+a_i)^2}\cdot p(x)=\sum_{i=1}^n \frac{a_i(a_i-\overline{a})(x+\overline{a})^2}{(x+a_i)^2}.$$

Existence of exactly one positive root of $q(x)$ follows from the next three statements:

(i) $q(0) < 0$;

(ii) $\lim\limits_{x\to+\infty} q(x) > 0$;

(iii) $q(x)$ is monotonically increasing, i.e., $q'(x)>0$ for all $x\geq 0$.


(i) It is easy to see that $$q(0) = \sum_{i=1}^n \frac{a_i(a_i-\overline{a})\overline{a}^2}{a_i^2} = \overline{a}^2\cdot \sum_{i=1}^n \left(1-\frac{\overline{a}}{a_i}\right)=\frac{\overline{a}^2}{n}\cdot \left( n^2 - \sum_{i=1}^n a_i\cdot\sum_{i=1}^n\frac{1}{a_i}\right).$$ We have $q(0)<0$ thanks to the Chebyshev sum inequality.

(ii) It is easy to see that $$\lim\limits_{x\to+\infty} q(x) = \sum_{i=1}^n a_i(a_i-\overline{a}) = \sum_{i=1}^n a_i^2 - n\overline{a}^2 = n\cdot \textrm{Var}(a_1,\dots,a_n)>0.$$

(iii) We have $$q'(x) = 2\cdot (x+\overline{a})\cdot \sum_{i=1}^n \frac{a_i(a_i-\overline{a})^2}{(x+a_i)^3},$$ which is positive for all $x\geq 0$, since all summands are nonnegative and cannot all be zero at the same time.


  • $\begingroup$ Neat derivation! $\endgroup$ – dima May 4 '15 at 12:58

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