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)



Positive roots of a polynomial
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](https://en.wikipedia.org/wiki/Chebyshev%27s_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


