Quasi-convexity of sum of two functions So I'm trying to prove the quasi-convexity of the following function:
$$f(x,y) = \frac{\sqrt{(a^Tx)^2 + (b^Tx)^2}}{y} - c^Tx = g(x,y) - c^Tx$$ 
where $x\in \mathbb{R}^n$ and $y \in \mathbb{R}_+$ and $c^Tx \in \mathbb{R}_+$. Now, I was able to prove that $g(x,y)$ is quasi-convex. It is known that the sum of two quasi-convex functions is not necessarily quasi-convex; however, is there a way to prove that the difference between a quasi-convex and positive linear function is quasi-convex?
Your help is much appreciated!
 A: The function you gave is, in general, not quasi-convex. Take the one-dimensional case - that is $x \in \mathbb{R}$. Taking $a = b = \tfrac{1}{\sqrt{2}},~ c = 1$, we get
$$
f(x,y) = \frac{|x|}{y} - x
$$
In our case $c^T x \in \mathbb{R}_+$ means $x > 0$. Thus, in this domain the function is
$$
f(x,y) = \frac{x}{y} - x
$$
This function is not quasi-convex. To prove it, assume the contrary. The $\alpha$ level set for $\alpha = \frac{1}{10}$ is
$$
\begin{aligned}
L 
  &= \{ (x,y) : \tfrac{x}{y} - x \leq \frac{1}{10}, ~x,y > 0 \} \\
  &= \{ (x,y) : 10x - 10xy - y \leq 0, ~x,y > 0 \}
\end{aligned}
$$
By quasi-convexity, $L$ is convex. Take $C = \{ (x,y) : y = 1.8x + 0.1 \}$. By properties of convex sets $L \cap C$ is convex. On the other hand
$$
L \cap C = \{ x: -180x^2 + 72x - 1 \leq 0, x > 0 \} = (0, \frac{6 - \sqrt{31}}{30}] \cup [\frac{6+\sqrt{31}}{30}, \infty),
$$
which is clearly convex. Thus, we got a contradiction, meaning that $f$ is not quasi-convex.
In addition, visually, $L$ is the white area in the plot below, which is clearly not convex.
A: is there a way to prove that the sum of a quasi-convex and linear function is quasi-convex?
No take $f(x) = x^3 -x$
