Prove the existence of answer: Let $K$ be closed and convex. also $F:K \subseteq \mathbb R^n \to \mathbb R^n $ be a continuous function.If foe every $x,y \in K$ we have $(x-y)^T(F(x)-F(y))\ge \alpha ||x-y||^2  ;\alpha\gt0 $ we want to prove that there exist a unique $x^*$ such that $F(x^*)^T(x-x^*)\ge 0$.
It is obvious proof of uniqueness of solutions.Assuming there are two answers we have $(x^*-x')^T(F(x')-F(x^*))\ge\alpha ||x-y||^2$ Which is a contradiction.
But how to prove that there exist answer?
 A: This is not true in general. If we take $F:\mathbb{R}\to \mathbb{R}$ to be $F(x)= 2+\arctan(x)$ then as $F$ is strictly increasing we have that $(x-y)(F(x)-F(y))>0$ for all $x\neq y\in\mathbb{R}$. However, for any $x\in\mathbb{R}$ we can just pick $y<x$ so that $F(x)(y-x)<0$.
I suspect that if we have the additional assumption that $K$ is compact, this may be true. However, a proof or counter example of this escapes me at the moment.
It may be worth noting that if we assume that $F$ is linear, then the first condition is exactly the $F$ is positive definite and this then becomes the usual characterization of the point in $K$ that is closest to the origin under the norm $||x||^2:=x^TF(x)$. 
Edit: Given your edit, here is a partial solution. We assume there exists a differentiable function $f:K\subseteq\mathbb{R}^n\to \mathbb{R}$ such that $\nabla f = F$. Notice that in this case, it is sufficient to prove that there exists a point $x^*\in K$ such that $f(x^*)$ is the global minimum as this then implies $(F(x^*),x-x^*)\geq 0$ since this is just the directional derivatives of $f$ at $x^*.$ The brackets above are just the usual inner product on $\mathbb{R}^n$.
Now, WLOG we will assume that $0\in K$. Let $x\in K$ and define $g(t):=f(tx)$ for $t\in [0,1]$ which is well defined due to convexity. 
Differentiating we get $$g^\prime(t)=(\nabla f(tx),x)$$ and by the FTC we have that $$ f(x)-f(0) = \int_0^1(\nabla f(tx),x)dt.$$
From the first condition we have that $$(\nabla f(tx)-\nabla f(0),tx)\geq\alpha ||tx||^2 $$ which implies that $$(\nabla f(tx),x) \geq \alpha t ||x||^2 + (\nabla f(0),x)\geq\alpha t ||x||^2 -||\nabla f(0)||\cdot ||x||. $$ Hence we have that $$f(x)-f(0)\geq \frac{\alpha}{2}||x||^2 -||\nabla f(0)||\cdot ||x||.$$
Since the RHS $\to \infty$ when $||x||\to \infty$ we conclude that there exists an $M>0$ large enough such that if $x\in K$ and $||x||>M$ then $f(x)>f(0)$. In particular, a minimum in $K\cap B(0,M)$ is a minimum in $K$, but $K\cap B(0,M)$ is compact and $f$ is continuous, hence $f$ achieves a global minimum.
