Let $f\colon S\to \mathbb R$ be a $C^1$ function on a convex domain $S \subseteq \mathbb R^n$. Show that if $f$ is convex then $(\nabla f(x) - \nabla f(y)) \cdot (x-y) \ge 0$ for all $x,y \in S$.
My intuition is that since $f$ is convex we have that $f(y) \ge f(x) + \nabla f(x)(y-x)$ and that $f(x) \ge f(y) +\nabla f(y)(x-y)$ and I'd like to use these somehow but I am not sure what the best approach would be. Is there a good geometric intuition for this statement, and any major ideas that might help me in tackling proofs of further properties?
Thanks!
If $\phi$ is a differentiable convex function of one variable then $\phi'(t) \geq \phi'(s)$ whenever $t \geq s$.
Now let $\phi(t) = f(y + t (x-y))$, compute $\phi$ and apply the above result to $\phi'(1)$ and $\phi'(0)$.