$\newcommand{\half}{\frac{1}{2}}$
Let $g:\mathbb R^2\to \mathbb R$ be a smooth symmetric function, i.e. $g(x,y)=g(y,x)$ for every $x,y$. Suppose that
$$
g\left( \half(x,y)+\half(y,x) \right) \le \half g(x,y)+\half g(y,x) =g(x,y), \tag{1}
$$
for every $x,y$ or equivalently
$$
g(x,y) \ge g(\frac{x + y}{2}, \frac{x + y}{2}) .
$$
Is $g$ midpoint convex?
That is, does $$ g(\frac{u + v}{2}) \le \frac{g(u) + g(v)}{2} \tag{2} $$ for every $u,v \in \mathbb R^2$?
Condition $(1)$ is more specific than $(2)$ since in inequality $(1)$, $u=(x,y),v=(y,x)$ are related to each other.
Counterexample: let $g(x, y) = (x - y)^2 + (x + y)^3$. Then $g$ is clearly smooth, and $g(x, y) = g(y, x)$. We have, $$g(x, y) - g\left(\frac{x + y}{2}, \frac{x + y}{2}\right) = (x - y)^2 + (x + y)^3 - (x + y)^3 \ge 0,$$ hence $g$ satisfies the premises of the question. But $g$ is not midpoint convex. If we take $u = (0, 0)$ and $v = (-2, -2)$, then, \begin{align*} g\left(\frac{u + v}{2}\right) &= g(-1, -1) = -8 \\ \frac{g(u) + g(v)}{2} &= \frac{0 - 64}{2} = -32 < g\left(\frac{u + v}{2}\right). \end{align*}