It is known that $L^1$ norm divided by $L^2$ norm is non-convex. I tried to prove it with the triangle inequality
$$\theta f(x) + (1-\theta)f(y) \ge f(\theta x+(1-\theta)y)$$
define $x$ to be vector $(a,b)$ $y$ to be vector $(c,d)$.
However, it is hard to tell the inequality will be violated as it is an inequality with $5$ variables. I manage to find $x$ and $y$ that will violate the inequality, but this will be much more difficult for higher dimension. What is the standard approach to prove this problem?
Since you already have a counterexample in low dimensions, you can easily construct a counterexample in higher dimensions by setting the other elements to $0$.
Let $f(x) = ||x||_1 / ||x||_2$. For $n=2$, a counterexample is $x=(2,1)$ and $y=(1,2)$, since $f(x) = f(y) = 3/\sqrt{5} \approx 1.3$ while $f((x+y)/2) = \sqrt{2} \approx 1.4$. For $n>2$ you just set the remaining elements to $0$. For example, for $n=5$, you take $x=(2,1,0,0,0)$ and $y=(1,2,0,0,0)$.