Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be differenciable and $[\nabla f(y)-\nabla f(x)]\cdot (y-x)\ge 0\ \forall x,y\in \mathbb{R}^n$. Is $f$ a convex function?
Since $f$ is differentiable then convexity is equivalent to the condition $$f(y)\ge f(x)+\nabla f(x)\cdot (y-x)\ \forall x,y\in \mathbb{R}^n\ \ \ \ \ \ (1)$$ So If there is a function whose derivative satisfy $$[\nabla f(y)-\nabla f(x)]\cdot (y-x)\ge 0\ \forall x,y\in \mathbb{R}^n\ \ \ \ \ \ (2)$$ but $(1)$ is not true then the statement is false.
I know that $$f(y)\ge f(x)+\nabla f(x)\cdot (y-x)\ \forall x,y\in \mathbb{R}^n\Rightarrow [\nabla f(y)-\nabla f(x)]\cdot (y-x)\ge 0\ \forall x,y\in \mathbb{R}^n,$$ but I can't prove the conversely or find a counterexample.
It turns out that the solution was very easy:
Since $[\nabla f(y)-\nabla f(x)]\cdot (y-x)\ge 0$, then,$$\nabla f(y)\cdot (y-x)\ge\nabla f(x)\cdot (y-x)=\nabla f(x+z)\cdot z\ge\nabla f(x)\cdot z,\ z=y-x.$$ Now, since $f$ is differentiable, then exist $c\in(0,1)$ such that $$f(x+z)=f(x)+\nabla f(x+cz)\cdot z=f(x)+\frac{1}{c}\nabla f(x+cz)\cdot cz$$ Now using $y=x+cz$ we have that $$f(x+cz)\cdot cz\ge f(x)\cdot cz$$ Then $$f(x+z)=f(x)+\frac{1}{c}\nabla f(x+cz)\cdot cz\ge f(x)+\frac{1}{c}\nabla f(x)\cdot cz=f(x)+\nabla f(x)\cdot z$$ Then if $w=x+z$, we have that $$f(w)\ge f(x)+\nabla f(x)\cdot (w-x)$$ So $f$ is convex.