Whether the following function is convex or not？ 
$f(x)=\frac{||Ax+b||_2^2}{c^Tx+d}$ on $\{x\in\Bbb R^n\ | \ c^Tx+d>0\}$, where $A\in\Bbb R^{m\times n}$, $b\in\Bbb R^m$, $c\in\Bbb R^n$ and $d\in\Bbb R$.

It seems like a linear fractional function, but not the same.
 A: It is enough to show that $h(y,z) = \|y\|^2 / z$ is convex over $\mathbb R^m \times \mathbb R_{++}$, since composition with affine transformations preserves convexity.
We also have $h(y,z) = \sum_i y_i^2 /z$, so it is enough to show that each component function is convex (addition preserves convexity).
Some further thought show that the problem reduces to proving $g(u,z) = u^2 / z$ is convex over $\mathbb R \times \mathbb R_{++}$. The epigraph of this function is
$$
\left\{ (u,z,t) : u^2 \le t z,\; z > 0 \right\} = \left\{ (u,z,t) : 
\begin{pmatrix}
t & u  \\
u & z 
\end{pmatrix} \succeq 0,\; z > 0 \right\}
$$
which is a convex set, since the PSD cone is convex. This proves the convexity of $g$.
A: To prove the convexity of $f(x)$ is equivalent to proving the convexity of $\mathrm{epi}f$, i.e., proving the convexity of set in $\mathbb{R}^{n+1}$:
\begin{equation}
\begin{aligned}
\mathrm{epi}f &=\{(x,\mu) \mid \mu \ge \frac{\|Ax+b\|_{2}^{2}}{c^{\mathsf{T}} x+d}, c^{\mathsf{T}} x + d > 0\} \\
&= \{(x,\mu) \mid \mu (c^{\mathsf{T}} x+d) \ge \|A x+b\|_{2}^{2}, c^{\mathsf{T}} x + d > 0\}
\end{aligned}
\end{equation}
Suppose $(x_1, \mu_1), (x_2, \mu_2) \in \mathrm{epi}~f$, then, for any $(x,\mu) = \lambda (x_1, \mu_1) + (1-\lambda)(x_2,\mu_2)$, where $0 \leq \lambda \leq 1$, we have
\begin{equation}
\begin{aligned}
(c^{\mathsf{T}}x+b)\mu = & \left(c^{\mathsf{T}}(\lambda x_1 + (1-\lambda)x_2)+d \right) \cdot (\lambda \mu_1 + (1-\lambda) \mu_2) \\
= &\lambda (c^{\mathsf{T}}x_1 + d) \cdot \lambda \mu_1 + (1-\lambda)(c^{\mathsf{T}}x_2 + d) \cdot (1-\lambda) \mu_2 \\
& +  \lambda (c^{\mathsf{T}}x_1  + d)\cdot (1-\lambda)\mu_2+ (1-\lambda)(c^{\mathsf{T}}x_2+d)\cdot \lambda \mu_1 \\
\ge &\lambda^2\|Ax_1+b\|_2^2 + (1-\lambda)^2 \|Ax_2+b\|_2^2 + 2\sqrt{\lambda^2(1-\lambda)^2\mu_1\mu_2 (c^{\mathsf{T}}x_1+d)(c^{\mathsf{T}}x_2+d)} \\
\ge & \lambda^2\|Ax_1+b\|_2^2 + (1-\lambda)^2 \|Ax_2+b\|_2^2 + 2\lambda(1-\lambda)\|Ax_1+b\| \cdot \|Ax_2+b\| \\
= & \|A (\lambda_1x_1 + \lambda_2 x_2) +b\|^2 = \|Ax+b\|^2.
\end{aligned}
\end{equation}
Thus $\mathrm{epi}f$ is convex set, so the function $f$ is convex function on its domain.
