convex-analysis
Let $\Omega$ be a set, e.g $\mathbb{R}^n$ for $n \in \mathbb{N}$.
Let $f$ be a convex function from $\Omega$ to $\mathbb{R}$.
Let $g$ be a quasi-convex function from $\Omega$ to $\mathbb{R}$.
Is $f+g$ quasi-convex? The proof is not straightforward from using the definition:
Let $\lambda \in \left[0,1\right]$. Let $(x,y) \in \Omega$. Then,
$\lambda (f+g)(x)+(1-\lambda)(f+g)(y) \leq max(g(x),g(y)) + \lambda f(x)+(1-\lambda)f(y)$.
which does not help to conclude anything.
The second is neither convex nor concave - that's easy to determine simply by looking at it. Along the line $y=x$, it is convex as a 1D function; along the line $y=-x$ it is concave. It is neither quasi-convex nor quasi-concave: to show not quasi-concave, consider the points $x = (0, 1)$, $y = (-1, 0)$, so $f(x) = f(y) = 0$. Parametrise the function along that line segment by $\lambda$; then $f(\lambda) = \lambda (\lambda - 1) < 0 = \min \{ f(x), f(y) \}$. You can rotate to get non-quasi-convexity.
The first is convex but not concave, and it's not quasi-concave. Examine the value of $f$ at the points $x=1/3, x=10, x=1$ to see that it's not quasi-concave. It's convex again by inspection or by showing that its second derivative is strictly positive.
This is false as stated. Quasiconvexity definition says that the lower level sets $\{f\le \lambda\}$ are convex. Unimodularity says that $f$ increases along the rays emanating from its global minimum. The first certainly implies the second. But the converse is false.
Example: $f(x_1,x_2) = \sqrt{|x_1|}+\sqrt{|x_2|}$. Clearly, $-f$ is unimodal with global maximum at $(0,0)$. But the lower level sets are not convex: they are bounded by astroids.
Concretely, $f(1/2,1/2) = \sqrt{2}$ is greater than $f(1,0)=f(0,1)=1$.
Best Answer
The statement is wrong for $\Omega = \mathbb{R}$.
Let $f(x)=-x$ and $g(x)=x-\frac{1}{2}|x|$. $f$ is obviously convex, and $g$ is monotonically increasing, and thus quasi-convex, but their sum $(f+g)(x)=-\frac{1}{2}|x|$ is obviously not quasi-convex.