Convex function which is strictly convex in one argument

Question

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a convex function which is strictly convex in its first argument, i.e. $x_1\mapsto f(x_1,x_2)$ is strictly convex for every $x_2\in\mathbb{R}$. Does it follow that for every $x,y\in\mathbb{R}^2$ with $x_1\neq y_1$ we have
$$
f(tx + (1-t)y) < tf(x) + (1-t)f(y)
$$
for every $t\in(0,1)$?

I am struggling to come up with a proof or a counterexample.

Answer 1

Using the idea of @CyclotomicField leads to $$ f(x_1, x_2) = (x_1 - x_2)^2 .$$