Solve $1/x^2 = \sum_{i=1}^{n} \frac{1}{x+a_i}$ over $x>0$ Does the equation 
$\frac{1}{x^2} = \sum_{i=1}^{n} \frac{1}{x+a_i}, \qquad a_i > 0, \quad i=1, \ldots, n$
always admits one and only one solution $x^* > 0$? If yes, what is the most elegant way to prove it?
 A: This is the same as the equation
$$1=\sum_{k=1}^{n}\frac{x^2}{x+a_k} = f(x).$$
Note $f$ is a nice $C^1$ function on $[0,\infty).$ Since
$$f'(x) =\sum_{k=1}^{n}\frac{x^2+2xa_k}{(x+a_k)^2}>0, x> 0,$$
$f$  is strictly increasing on $[0,\infty).$ We have $f(0) = 0,$ and $f\to \infty$ at $\infty,$ so the intermediate value theorem shows $f$ takes on each value in $[0,\infty).$ Because $f$ is strictly increasing, each of these values is taken on exactly once, in particular the value $1.$
A: For $x > 0$, the equation can be rewritten as
$$\frac{1}{x} = \sum_{i=1}^n \frac{x}{x+a_i} = \sum_{i=1}^n\left(1 - \frac{a_i}{x+a_i}\right)\quad\iff\quad
\frac{1}{x} + \sum_{i=1}^n \frac{a_i}{x+a_i} = n$$
Since the LHS of the expression on the right is a strictly decreasing in $x$, the equation has at most one positive $x$ solution. 
In fact, since that LHS is a continuous function $x$ which blows up to $+\infty$ for $x \to 0$ and tends to $0$ as $x \to +\infty$, the equation has an unique positive $x$ solution.
