This isn't quite what you want, but....
On the 1971 Putnam, there was a question, show that if $n^c$ is an integer for $n=2,3,4,\dots$ then $c$ is an integer.
If you try to improve on this by proving that if $2^c$, $3^c$, and $5^c$ are integers then $c$ is an integer, you find that the proof depends on a very deep result called The Six Exponentials Theorem.
And if you try to improve further by showing that if $2^c$ and $3^c$ are integers then $c$ is an integer, well, that's generally believed to be true, but it hadn't been proved in 1971, and I think it's still unproved.
If you have some data $y$ and want to fit it to a curve of the form $1/(ax^2+bx+c)$, then what we can do is fit the data $1/y$ to the curve $ax^2+bx+c$. To do this in the least squares sense, we construct the following system of equations:
$$ \begin{pmatrix} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ \vdots & \vdots & \vdots \\ x_n^2 & x_n & 1\end{pmatrix} \begin{pmatrix} a \\ b \\ c\end{pmatrix} =
\begin{pmatrix}
1/y_1 \\ 1/y_2 \\ \vdots \\ 1/y_n
\end{pmatrix}
$$
The matix on the left-hand side is known as a Vandermonde matrix. Denote this matrix $A$
Since in general $n \neq 3$, to solve this we simply solve $A^TA \mathbf{p} = A^T\mathbf{y}$, where $\mathbf{p} = (a,b,c)^T$ is the parameter vector.
Thus, $$\mathbf{p} = (A^TA)^{-1}A^T\mathbf{y},$$ which may be computed in many ways (note that $A^TA$ is symmetric, so the LU factorization should be easy to compute).
EDIT: Regarding largeness of $\mathbf{y}$ (which should not matter for the data in your picture).
Note that we are solving $A^TA\mathbf{p} = A^T\mathbf{y}$. Look at $A^T\mathbf{y}$:
$$A^T \mathbf{y} = \begin{pmatrix}
\sum_{i=1}^n \frac{x_i^2}{y_1} \\
\sum_{i=1}^n \frac{x_i}{y_1} \\
\sum_{i=1}^n \frac{1}{y_1}
\end{pmatrix}$$
If $y_i$ is so small that you are concerned about numerical stability, solve the following problem instead:
$$10^m \hat{a} x^2 + 10^m \hat{b} x + 10^m \hat{c} = 1/y$$
for some $m > 0$. Then, you condition your Vandermonde matrix in such a way that $A^T\mathbf{y}$ becomes:
$$A^T \mathbf{y} = \begin{pmatrix}
\sum_{i=1}^n \frac{10^m x_i^2}{y_1} \\
\sum_{i=1}^n \frac{10^m x_i}{y_1} \\
\sum_{i=1}^n \frac{10^m}{y_1}
\end{pmatrix}$$
and the smallness is gone. Compute $\hat{a},\hat{b},\hat{c}$ in the traditional sense, then $a = 10^m\hat{a}$, and so forth.
Best Answer
This is probably because of following theorem on positive polynomials.
Please note that the corresponding statement is false for polynomials in two or more variables. e.g. the Motzkin polynomial
$$ x^4y^2 + x^2y^4 - 3x^2y^2 + 1$$
is non-negative over $\mathbb{R}^2$ and yet it cannot be expressed as a sum of two real polynomials.
The question whether any non-negative real polynomials in $n > 1$ variables can be expressed as a sum of squares of real rational functions is the famous Hilbert's seventeen problem. Follow the wiki links for more details and inspirations.