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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

UPDATE. The same statement holds if in the formula for $p(x)$ second power is replaced by any other positive power.

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

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

UPDATE. The same statement holds if in the formula for $p(x)$ second power is replaced by any other positive power.

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

UPDATE. The same statement holds if in the formula for $p(x)$ second power is replaced by any other positive power.

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 $$q(x):=\frac{(x+\overline{a})^2}{\sum_{i=1}^n (x+a_i)^2}\cdot p(x)=\sum_{i=1}^n \frac{a_i(a_i-\overline{a})(x+\overline{a})^2}{(x+a_i)^2}.$$

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

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

UPDATE. The same statement holds if in the formula for $p(x)$ second power is replaced by any other positive power.