How to solve $\mathrm{diag}(x) \; A \; x = \mathbf{1}$ for $x\in\mathbb{R}^n$ with $A\in\mathbb{R}^{n \times n}$? I would like to solve the following equation for $x\in\mathbb{R}^{n}$
$$\mathrm{diag}(x) \; A \; x = \mathbf{1}, \quad \text{with $A\in\mathbb{R}^{n\times n}$},$$
where $\mathrm{diag}(x)$ is a diagonal matrix whose diagonal elements are the elements of $x$ and $\mathbf{1}$ is a vector whose elements are equal to 1.
I will already be very happy to find a solution if $A$ is a positive definite and symmetric.
Ideally I would like to find a closed-form solution for this quadratic equation.
Any ideas (or solution ;-) would be greatly appreciated.
Other formulation
Another way to formulate this equation is as follows
$$A \; x = 1./x, \quad \text{with $A\in\mathbb{R}^{n\times n}$},$$
where $1./x$ denotes the "element-wise inverse of the vector $x$".
Solution for the 1-dimensional case
The solution for the 1-dimensional case is straightforward
$$ x = \frac{1}{\sqrt{A}} .$$

Do these least squares solutions help?

$n=2$

$$ 
\begin{align}
%
\mathbf{D} \mathbf{A} x &= \mathbf{1} \\
%
\left[
\begin{array}{cc}
 a_{\{1,1\}} & a_{\{1,2\}} \\
 a_{\{2,1\}} & a_{\{2,2\}} \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
 x_{\{1\}} & 0 \\
 0 & x_{\{2\}} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x_{\{1\}}\\
 x_{\{2\}} \\
\end{array}
\right]
%
&= 
%
\left[
\begin{array}{c}
 1 \\
 1 \\
\end{array}
\right]
%
\end{align}
$$
The least squares solution is
$$ 
\begin{align}
%
 x_{LS} &= \left( \mathbf{D} \mathbf{A} \right)^{+} \mathbf{1} \\
%
&= \left( \det \mathbf{DA} \right)^{-1}
\left[
\begin{array}{c}
\frac{a_{\{2,2\}}}{x_{\{1\}}}-\frac{a_{\{1,2\}}}{x_{\{2\}}} \\
\frac{a_{\{1,1\}}}{x_{\{2\}}}-\frac{a_{\{2,1\}}}{x_{\{1\}}}\end{array}
\right]
%
\end{align}
$$

$n=3$

$$ 
\begin{align}
%
\mathbf{D} \mathbf{A} x &= \mathbf{1} \\
%
\left[
\begin{array}{ccc}
 x_{\{1\}} & 0 & 0 \\
 0 & x_{\{2\}} & 0 \\
 0 & 0 & x_{\{3\}} \\
\end{array}
\right]
%
\left[
\begin{array}{ccc}
 a_{\{1,1\}} & a_{\{1,2\}} & a_{\{1,3\}} \\
 a_{\{2,1\}} & a_{\{2,2\}} & a_{\{2,3\}} \\
 a_{\{3,1\}} & a_{\{3,2\}} & a_{\{3,3\}} \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x_{\{1\}} \\
 x_{\{2\}} \\
 x_{\{3\}} \\
\end{array}
\right]
%
&= 
%
\left[
\begin{array}{c}
 1 \\
 1 \\
\end{array}
\right]
%
\end{align}
$$
The least squares solution is
$$ 
\begin{align}
%
 x_{LS} &= \left( \mathbf{D} \mathbf{A} \right)^{+} \mathbf{1} \\
%
&= \left( \det \mathbf{DA} \right)^{-1}
\left[
\begin{array}{c}
%
\frac{a_{\{1,3\}} a_{\{3,2\}}-a_{\{1,2\}} a_{\{3,3\}}}{x_{\{2\}}}+a_{\{2,3\}} \left(\frac{a_{\{1,2\}}}{x_{\{3\}}}-\frac{a_{\{3,2\}}}{x_{\{1\}}}\right)+a_{\{2,2\}} \left(\frac{a_{\{3,3\}}}{x_{\{1\}}}-\frac{a_{\{1,3\}}}{x_{\{3\}}}\right) \\
%
\frac{a_{\{1,1\}} a_{\{3,3\}}-a_{\{1,3\}} a_{\{3,1\}}}{x_{\{2\}}}+a_{\{2,1\}} \left(\frac{a_{\{1,3\}}}{x_{\{3\}}}-\frac{a_{\{3,3\}}}{x_{\{1\}}}\right)+a_{\{2,3\}} \left(\frac{a_{\{3,1\}}}{x_{\{1\}}}-\frac{a_{\{1,1\}}}{x_{\{3\}}}\right) \\
%
\frac{a_{\{1,2\}} a_{\{3,1\}}-a_{\{1,1\}} a_{\{3,2\}}}{x_{\{2\}}}+a_{\{2,2\}} \left(\frac{a_{\{1,1\}}}{x_{\{3\}}}-\frac{a_{\{3,1\}}}{x_{\{1\}}}\right)+a_{\{2,1\}} \left(\frac{a_{\{3,2\}}}{x_{\{1\}}}-\frac{a_{\{1,2\}}}{x_{\{3\}}}\right)
%
\end{array}
\right]
%
\end{align}
$$
