Positive roots of a polynomial 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)
 A: 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 $x\geq 0$.
Proof.
(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.
QED
