TITLE: Proving that the roots of $1/(x + a_1) + 1/(x+a_2) + ... + 1/(x+a_n) = 1/x$ are all real
QUESTION [8 upvotes]: Prove that the roots of the equation:
$$\frac1{x + a_1} + \frac1{x+a_2} + \cdots + \frac1{x+a_n} = \frac1x$$
are all real, where $a_1, a_2, \ldots, a_n$ are all negative real numbers.

REPLY [2 votes]: HINT:
$$\dfrac1x=\sum_{r=1}^n\dfrac1{x+a_r}\iff n=\sum_{r=1}^n\dfrac{a_r}{x+a_r}$$
Let $p+iq$ is a root where $p,q$ are real
$$\implies n=\sum_{r=1}^n\dfrac1{p+iq+a_r}$$
Using  Complex conjugate root theorem, $$n=\sum_{r=1}^n\dfrac1{p-iq+a_r}$$
Subtracting we get $$n-n=\sum_{r=1}^n\left(\dfrac1{p-iq+a_r}-\dfrac1{p+iq+a_r}\right)$$
$$\iff0=2iq\sum_{r=1}^n\dfrac1{(p+a_r)^2+q^2}$$