Proving that the roots of $1/(x + a_1) + 1/(x+a_2) + ... + 1/(x+a_n) = 1/x$ are all real

We can prove a stronger statement: the equation above has $n - 1$ real positive roots and a negative real one, and there are no other roots.

Let $$g(x) = \sum_{i = 1}^n \frac1{x-a_i} - \frac1x,\qquad a_i \in (0, +\infty).$$

Note that $g(x)$ is defined in $\mathbb R \setminus \{0, a_1, \ldots, a_n\}$ and it's also continuous. Without loss of generality, suppose that $a_1 < a_2 < \cdots < a_n$.
Now, consider the interval $(a_i, a_{i + 1})$. We have that:

• $\lim\limits_{x \to a_i^+} g(x) = +\infty$
• $\lim\limits_{x \to a_{i + 1}^-} g(x) = -\infty$

Therefore, from the definition of limit and the intermediate value theorem, we deduce that there is a root in $(a_i, a_{i + 1})$. We proved the existence of $n - 1$ real positive roots.

It can be easily verified that

• $\lim\limits_{x \to -\infty} g(x) = 0^-$
• $\lim\limits_{x \to 0^-} g(x) = +\infty$

Again, from the definition of limit and the intermediate value theorem we conclude that there is another real root in the interval $(-\infty, 0)$.

We are done: observe that the equation $g(x) = 0$ is equivalent, through some simple algebra, to a polynomial equation of degree $n$. Having found $n$ real roots, we conclude that there are no complex roots.

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 $$n-n=\sum_{r=1}^n\left(\dfrac1{p-iq+a_r}-\dfrac1{p+iq+a_r}\right)$$

$$\iff0=2iq\sum_{r=1}^n\dfrac1{(p+a_r)^2+q^2}$$