Suppose $F(x+y) \geq \frac{x}{x+y} F(x) + \frac{y}{x+y}F(y)$ for all $x ,y \in \mathbb{R}_+$. Is $F$ convex? Let $F : \mathbb{R}_+ \to \mathbb{R}$ be a function which is controlled in the following way: $$F(x+y) \geq \frac{x}{x+y} F(x) + \frac{y}{x+y} F(y).$$ Is this function convex?
This estimate is satisfied by convex functions, i.e., $F(x) = e^x$ and $F(x)=x^2$: \begin{eqnarray*}
(x+y) e^{x+y} &=& xe^{x+y} + y e^{x+y} \ \geq \ x e^x + ye^y, \\
(x+y)(x+y)^2 & =& x \cdot x^2 + y \cdot y^2 + (\text{positive cross terms})
\end{eqnarray*}
It might be worth noting that if $t = \frac{x}{x+y} \in (0,1]$, then the defining estimate reads: $$F\left( \frac{x}{t} \right) \geq t F(x) + (1-t) F \left( \frac{x}{t} -1 \right).$$

$f$ does not have to be convex.
Try for example $f(x)=-x^{-½}$, for $x>0$, which is concave and satisfies
$$
f(x+y)>\frac{x}{x+y}f(y)+\frac{x}{x+y}f(y)
$$
