If $f(x)=x$ for $x \in (0,2]$ and $f(x)=\frac{x+2}{2}$ for $x>2$ then $f(x)$ is strictly increasing and $f(2)=\frac{f(1)+f(4)}{2}$ but $f(x)>0$ for every $x>0$

The problem condition probably means "for all $x$, $y$" and not "for some $x$, $y$".

Nice. Assume the contrary that $f(x)\ge 0$ for all $x>0$. Take $x=t^2$ and $y=1$ to find that $f(t)\ge f(t^2)/2+f(1)/2\ge f(1)/2$ for all $t$. We now construct a sequence $a_n$ such that $a_1=1/2$, $a_{n+1}=1/2+a_n/2$ and $f(t)\ge f(1)a_n$ for all $t$ and $n$. Note that once $f(t)\ge f(1)a_n$, we get that $f(t^2)\ge f(1)a_n$, too; yielding $f(t)\ge f(1)(1/2 + a_n/2) = f(1)a_{n+1}$.Clearly $a_n\to 1$. As a result, we recover $f(t)\ge f(1)$ after sending $n\to\infty$. This is a clear contradiction if $t<1$.

I think we can just solve to find $f$. Let $x=e^u,y=e^v$ with $u,v \in \mathbb{R}$ we have $f\left(e^{\frac{x+y}{2}} \right)=\dfrac{1}{2}\left( f(e^x)+f(e^y) \right)$ Let $g(x)=f(e^x)$, we have $g\left(\dfrac{x+y}{2} \right)=\dfrac{1}{2} \left(g(x)+g(y) \right) \forall x,y \in \mathbb{R}$ (1) Note that if $g(x)$ is satisfied, $g(x)+a$ is also satisfied, so we can assume $g(0)=0$ Let $y=0$ in (1) we have $g\left(\dfrac{x}{2} \right)=\dfrac{1}{2} g(x)$ or $g(2x)=2g(x) \forall x \in \mathbb{R}$ Plug in the condition we have $g(x)+g(y)=2g\left(\dfrac{x+y}{2} \right)=g(x+y)$ For all $x>y$ we have $g(x)=f(e^x)>f(e^y)=g(y)$, therefore $g$ is strictly increasing. So, $g(x)=ax+b$ with $a>0$, or $f(x)=\ln (ax+b)$ Plugging in the condition we have $f(x)=\ln (ax)$ or $f(x)=k\ln x$ with $k>0$ satisfied.