Convexity and Superadditivity in $\mathbb{R}^n$

Question

In 1-dimension convex function are superadditive on $\mathbb{R}^+$ with $f(0) \leq 0$ (easy proof)

How to prove superadditivity of convex function in $\mathbb{R}^n$ for $n \geq 2$? (if possible?)

Convex function : For $\lambda \in [0,1]$ and $a, b \in \mathbb{R}^n$ if f satisfies $f(\lambda a +(1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$

Answer 1

It's not even true that a convex function is superadditive on $\Bbb R$. Consider the function $f(x)=x^2$, we have $$ f(1+(-1)) = 0 < 2 = f(1) + f(-1). $$