Let f be a function such that f is Frechet differentiable. Prove that f is concave if and only if the following inequality holds: $$ 0\le f(x)+f(y)+\left<x,\nabla(f)(y)\right>+\left<y,\nabla(f)(x)\right>,\forall x,y\in{\mathbb{R}^n} $$ where $\left<\cdot,\cdot\right>$ denotes the dot product between two vectors in ${\mathbb{R}^n}$
My progress:
I immediately thought that the inequality in my problem relates to the following fairly well known lemma: $f$ is concave iff $$ f(y)\le f(x) + \left<\nabla(f)(x),y-x\right>,\forall x,y\in{\mathbb{R}^n} $$ I can prove this by using the definition of concavity, so there is no need to dwell here.
I am stuck on how exactly to use the above result in my proof. I thought of doing this, but I am not sure if it's right: $$ f(y)\le f(x) + \left<\nabla(f)(x),y-x\right>,\forall x,y\in{\mathbb{R}^n} $$ $$ f(x)\le f(y) + \left<\nabla(f)(y),x-y\right>,\forall x,y\in{\mathbb{R}^n} $$ Now sum the two and get $$ 0\le\left<\nabla(f)(x),y-x\right>+\left<\nabla(f)(y),x-y\right> $$ Is any of this good? Do you have alternate suggestions on how to prove the initial statement? EDIT: Forgot $f(y)$ in my initial statement. Many apologies.
It cannot be proven, since it is false. Take $f(x) = -x^2$ so that $f$ is concave. Then your first equation states that $0 \le -x^2 -4xy$ for all $x,y \in \mathbb{R}$.