Just a comment, not an answer, but perhaps worth a try.

Assuming $x \not= -\alpha_1,\ldots,-\alpha_n$, we can rewrite the equation as
\begin{equation*}
\prod_j (1+x\alpha_j)\sum_i \frac{(1-x^2\alpha_i)}{1+x\alpha_i} = 0.
\end{equation*}
Under our assumption, this further simplifies to the OP's [older equation:][1]
\begin{equation*}
f(x) := \sum_i \frac{1-x^2\alpha_i}{1+x\alpha_i}=0.
\end{equation*}
Perhaps at this point, you could try to: (i) bracket the root, obtaining points $a<x^*<b$ such that $f(a)<0$, $f(x^*)=0$ and $f(b)>0$, where the bounds are reasonable functions of the $\alpha_i$; (ii) run (by hand) an iteration or two of Newton's method to get an approximate $x^*$ (seems hard); (iii) try [Lagrange inversion][2]; or (iv) try a fixed-point iteration.

---
*Example of (iv)* Rewrite the equation as follows
\begin{equation*}
x^2\sum_i \frac{\alpha_i}{1+x \alpha_i} = \sum_i \frac{1}{1+x\alpha_i}.
\end{equation*}
Using the notation $a(x)=\sum_i \alpha_i/(1+x \alpha_i)$ and $b(x)=\sum_i 1/(1+x \alpha_i)$, we then "solve" (to be proved) for $x$ by running the iteration
\begin{equation*}
 x_{k+1} = [b(x_k)/a(x_k)]^{1/2},\quad k=0,1,\ldots.
\end{equation*}
An experiment suggests that started from $x_0=0$, this iteration monotonically increases $x$ (*proof?*). Since $b(x)/a(x)$ is bounded above, the sequence $(x_k)$ must converge. Using a couple of steps of this iteration, one can get a "reasonable" approximation to the the desired solution $x^*$.


  [1]: http://mathoverflow.net/q/259875/8430
  [2]: https://en.wikipedia.org/wiki/Lagrange_inversion_theorem