Convexity of $x \mapsto \frac{\|Ax-b\|^2_2}{1-\|x\|^2_2}$ 
Given matrix $A \in \Bbb R^{n \times n}$ and vector $b \in \Bbb R^n$, determine whether the following function is convex.
$$ f: \{x \in \Bbb R^n : \|x\|_2 < 1\} \to \Bbb R, \qquad x \mapsto \frac{\|Ax-b\|^2_2}{1-\|x\|^2_2} $$


The numerator appears to be similar to the Least square function which I know is convex. Since: for convex function f over a convex set for any α ≥ 0, the function α*f is a convex function.
Is setting:
$a=\frac{1}{1-\|x\|^2_2} \geq 0$
a valid proof for convexity? I am very thankful for any tips!
 A: The function is convex. Indeed, clearly $S = \{x\in \mathbb{R}^n: \|x\|_2 < 1\}$
is a convex set, $\|Ax-b\|_2$ is convex and nonnegative on $S$, 
$1-\|x\|_2^2$ is concave and positive on $S$ (since $\|x\|_2^2$ is convex on $\mathbb{R}^n$), 
from Fact 1 below, the desired result follows.
Fact 1: Let $S\subseteq \mathbb{R}^n$ be a convex set. 
Let $f: S \to \mathbb{R}$ be convex and $f(x) \ge 0, \forall x \in S$. Let $g: x \to \mathbb{R}$ be concave and $g(x) > 0, \forall x \in S$.
Then $\frac{f^2}{g}$ is convex on $S$.
Proof of Fact 1: For any $x_1, x_2 \in S$ and $t\in [0,1]$, we have
$$f(t x_1 + (1-t)x_2) \le tf(x_1) + (1-t)f(x_2)$$
and
$$g(tx_1 + (1-t)x_2) \ge tg(x_1) + (1-t)g(x_2).$$
Thus, we have
\begin{align}
&t\frac{(f(x_1))^2}{g(x_1)} + (1-t)\frac{(f(x_2))^2}{g(x_2)}
- \frac{(f(tx_1 + (1-t)x_2))^2}{g(tx_1 + (1-t)x_2)}\\
\ge\ & t\frac{(f(x_1))^2}{g(x_1)} + (1-t)\frac{(f(x_2))^2}{g(x_2)}
- \frac{(tf(x_1) + (1-t)f(x_2))^2}{tg(x_1) + (1-t)g(x_2)}\\
=\ & \frac{t(1-t)[f(x_1)g(x_2) - f(x_2)g(x_1)]^2}{g(x_1)g(x_2)[tg(x_1) + (1-t)g(x_2)]}\\
\ge\ & 0.
\end{align}
Q.E.D.
