$(\nabla f(x) - \nabla f(y))^T (x-y) \geq 0$
$f$ be twice differentiable, with $\mathrm{dom}(f)$ convex.
I can derive the convexity of $f$ from the above inequality through first order condition, but is that also the sufficiency for convexity of $f$?
This is something that can be found in textbooks, like the Nesterov book Introductory lectures on convex optimization. You can fix $x$ and $y$ in $\mathbb{R}^n$ and define $h(t) = f(x+t(y-x))$ for $t \in \mathbb{R}$. Then $h(1) = h(0) + \int_0^1h'(t)dt$ and so \begin{align} \underbrace{f(y)}_{h(1)} &= \underbrace{f(x)}_{h(0)} + \int_0^1 \underbrace{\nabla f(x+t(y-x))^T(y-x)}_{h'(t)}dt\\ &= f(x) + \nabla f(x)^T (y-x) + \int_0^1 [\nabla f(x+t(y-x)) -\nabla f(x)]^T(y-x)dt\\ &=f(x) + \nabla f(x)^T (y-x) + \int_0^1 \frac{1}{t}\underbrace{[\nabla f(x+t(y-x)) -\nabla f(x)]^T(t(y-x))}_{\geq 0}dt\\ \end{align}