$(x^2+y^2)^{\frac{1}{4}}$

For your new question: functions that satisfy your inequality don't exist

Proof.

Suppose $f(x)+f(x+u+v)> f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$. Let us get a contradiction from it. Take any square insribed in a circle, and rotate it leaving insrcibed. Rotating continuously for the angle 90 degrees you can exchange to pairs of opposite vertices.

Here is the previous counterexample:

$(x^2+y^2)^{\frac{1}{4}}$

$F(x,y)=1$ will do

$(x^2+y^2)^{\frac{1}{4}}$