Suppose that, for each $x,y$, such that $f(x)\geq f(y)$ one of the following condition must hold:
$f(ax+(1-a)y)\geq f(y)$ for all $a\in(0,1)$
$f(ax+(1-a)y)\leq f(x)$ for all $a\in(0,1)$
$f$ is smooth. Can we claim that $f$ is either a quasi-convex function or a quasi-concave function?
Basically, we want to show that, either (1) holds for all pair $x,y$, or (2) holds for all pair $x,y$.
The domain of $f$ is $\mathbb R^N$ but I think we could try $N=1$ or $2$ first.
I try to find the set $S=\{x||f(x)=f(y)\}$. The boundary is smooth. Suppose that property (1) hold for two points $a,b\in S$, then property (1) must hold for an open neighborhood of $a,b$. Since $f$ is smooth, the neighborhood of neighborhood can cover all points of $S$ if $S$ is connected. This can be done because, in my humble opinion, if $f$ is smooth, then $S$ must be closed.
Is this approach plausible?
This approach is not possible. If $A$ is an open set, then $A$ is also a neighborhood of $A$. Thus one cannot hope to reach larger sets just by "taking neighborhoods".