Let $(X,d)$ be a complete metric space, let $F: X\rightarrow X$ such that $$\exists L > 1, \forall (x,y)\in X^2, d(F(x),F(y))>L\cdot d(x,y).$$ Show that if $F(X)=X$ then there is exactly one point $x_0\in X$ such that $F(x_0)=x_0$
I completely don't know how to bite that because it's the exact opposition of the Banach theorem. I would appreciate any hints.
In this case $F$ is injective : if $F(x) = F(y)$ then $0 = d(F(x),F(y) \geq L d(x,y)$ implies that $d(x,y) = 0$, that is, that $x=y$. As $F$ is surjective, it is bijective, and its inverse $F^{-1}$ is Lipschitz with coefficient $1/L < 1$, so as $X$ is complete, this inverse $F^{-1}$ has a fixed point $x_0$, which is also a fixed point for $F$. Finally, this $x_0$ is unique, as $F$ is injective.