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

Question

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.

Answer 1

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.

Answer 2

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

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

Answer 3

Your question is solved in

Number of real roots of $\frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+...+\frac{a_n}{a_n-x}=2016$ for $0<a_1<...<a_n$?

because

$1=\sum\limits_{k=1}^n \frac{x}{x+a_k}=\sum\limits_{k=1}^n \frac{1/|a_k|}{1/|a_k|-1/x}$ .

Answer 4

This is a fairy intuitive way of approaching it.

Consider 2 functions, $$y=\frac{1}{x}$$ $$y=\sum_{i=1}^n \frac{1}{x+a_i}$$ Since $a_i>0$, graph the 2 functions. Graph

Without loss of generality, assume that $\prod_{i=1}^n a_i\neq0 $

Trivially, the original equation is $n-1$th-degree polynomial equation, so by the Fundamental Theorem of Algebra, it can have at most $n-1$ roots.

From the graph, we can identify $n-1$ intersection points.

Therefore, all roots are real.

Answer 5

It is not very convenient to prove this assertion for negative numbers. Let us assume $$ 0<a_1<a_2<\cdots<a_n $$ and consider the function $$ f(x)=\frac{1}{x-a_1}+\ldots+\frac{1}{x-a_n}-\frac{1}{x} $$ $f(x)$ is continues at $(-\infty,0)$, $(0,a_1)$, $(a_1,a_2)$ , $\ldots$ , $(a_{n-1},a_n)$ , $(a_n,\infty)$.

Note that if $x=-\sum_{i=1}^{n} a_i$, then $f(x)<0$ (why?). In addition, $$ \lim_{x\to0^{-}}f(x)=\infty $$ so by the Intermediate value theorem there exist $x_1\in(-\infty,0)$ such that $f(x_1)=0$. Similarly, $$ \lim_{x\to a_1^{+}}f(x)=\infty\qquad\lim_{x\to a_2^{-}}f(x)=-\infty $$ so again by the Intermediate value theorem there exist $x_2\in(a_1,a_2)$ such that $f(x_2)=0$. In the same reasoning we can prove that there exist $x_i\in(a_{i-1},a_i)$ such that $f(x_i)=0$ for each $2\leq i\leq n$. We have found at least $n$ solutions for $f(x)=0$. Since $$ f(x)=\frac{\text{polynomial of $n$ degree}}{(x-a_1)\cdots(x-a_n)x} $$ We deduce that $f(x)=0$ have at most $n$ solutions. Hence $f(x)=0$ have exactly $n$ solutions, as desired.