Is it true that for any $n\geq 1$, any $a_1,\cdots,a_n \in \mathbb{R}_{>0}$ and any $b_1,\cdots, b_n \in \mathbb{R}$, the function:$$f(x) = \frac{b_1}{x+a_1} + \frac{b_2}{x+a_2} + \cdots + \frac{b_n}{x+a_n}$$always have at most 1 positive real root, i.e. at most one $x^* \in \mathbb{R}_{>0}$ such that $f(x^*) = 0$?
I ran some experiments on Mathematica, here's an example output for some randomized $a_k, b_k, k=1,\cdots,n$:Image may be NSFW.
Clik here to view.
The graph always shows 1 positive root, or no positive root at all. I also checked the polynomial $g(x) := \prod_{i=1}^n(x+a_i)f(x)$, and it looks like the sign of the coefficients always change at most once. This is consistent with the descartes rule of signs.
Could anyone help me prove the claim, or provide me a counter-example?
In fact, what I really need for what I'm doing is the following special case: Is it true that for any $n \geq 1, n' \geq 1$, any $a_1,\cdots,a_n, a'_1,\cdots, a'_{n'} \in \mathbb{R}_{>0}$, and $C\geq 1$, the function$$f(x) = \frac{1}{x+a_1} + \cdots + \frac{1}{x+a_n} - C\left(\frac{1}{x+a'_1} + \cdots + \frac{1}{x+a'_{n'}}\right)$$always have at most 1 positive real root? But I thought it would also be interesting to investigate the more general statement.
