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Not a full answer, but some comments and an algorithm that the OP may find relevant.

Assuming $x \not= -1/\alpha_1,\ldots,-1/\alpha_n$, 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: \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; 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 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 $x$ (some calculation required, but one can show this easily). 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 $x^*$ (this seems to converge quite fast, so could be a reasonable method to solve for $x$ numerically too).

EDIT: It seems that this Fixed-Point Iteration (FPI) is used in the preprint authored by the OP (albeit, without attribution)!

Not a full answer, but some comments and an algorithm that the OP may find relevant.

Assuming $x \not= -1/\alpha_1,\ldots,-1/\alpha_n$, 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: \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; 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 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 $x$ (some calculation required, but one can show this easily). 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 $x^*$ (this seems to converge quite fast, so could be a reasonable method to solve for $x$ numerically too).

EDIT: It seems that this Fixed-Point Iteration (FPI) is used in the preprint authored by the OP (albeit, without attribution)!

EDIT 2: Meanwhile (as noted in the comments below) the OP has fixed the attribution; please check the link the his comment for the latest version.

Not a full answer, but some comments and an algorithm that the OP may find relevant.

Assuming $x \not= -1/\alpha_1,\ldots,-1/\alpha_n$, 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: \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; 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 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 $x$ (some calculation required, but one can show this easily). 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 $x^*$ (this seems to converge quite fast, so could be a reasonable method to solve for $x$ numerically too).

EDIT: It seems that this FPI is used in the preprint (albeit, without attribution).

Not a full answer, but some comments and an algorithm that the OP may find relevant.

Assuming $x \not= -1/\alpha_1,\ldots,-1/\alpha_n$, 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: \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; 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 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 $x$ (some calculation required, but one can show this easily). 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 $x^*$ (this seems to converge quite fast, so could be a reasonable method to solve for $x$ numerically too).

EDIT: It seems that this Fixed-Point Iteration (FPI) is used in the preprint authored by the OP (albeit, without attribution)!

Not a full answer, but some comments and an algorithm that the OP may find relevant.

Assuming $x \not= -1/\alpha_1,\ldots,-1/\alpha_n$, 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: \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; 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 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 $x$ (some calculation required, but one can show this easily). 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 $x^*$ (this seems to converge quite fast, so could be a reasonable method to solve for $x$ numerically too).

Not a full answer, but some comments and an algorithm that the OP may find relevant.

Assuming $x \not= -1/\alpha_1,\ldots,-1/\alpha_n$, 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: \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; 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 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 $x$ (some calculation required, but one can show this easily). 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 $x^*$ (this seems to converge quite fast, so could be a reasonable method to solve for $x$ numerically too).

EDIT: It seems that this FPI is used in the preprint (albeit, without attribution).