$f \colon \mathbb{R}^n \to \mathbb{R}^n$ of class $C^1$ such that $||f(x) - f(y)|| \geq c||x-y||$ is a diffeomorphism. 
Let $f \colon \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map. Suppose that exists $c>0$ such that $||f(x)-f(y)||\geq c||x-y||$ for all $x,y \in \mathbb{R}^n$. Prove that $f$ is a diffeomorphism.

I know prove that $f \colon X \to X$, $f\in C^{0}$ such that $||f(x)-f(y)||\geq c||x-y||$ is a homeomorphism if $X$ is compact. It is a long and 
wearisome proof.
About the question, we have automatically that $f$ is injective, and it is open, cause $f$ is continous and injective, so we can consider a inverse, $h\colon f(\mathbb{R}^n) \to \mathbb{R}^n$, and it is continuous because $f$ is open. $A \subset \mathbb{R}^n$ open set $\implies$ $h^{-1}(A) = f(A)$ open.
My two difficulties are show that $f$ is surjective and $h$ is differentiable.
To show that $h$ is differentiable I was trying show that $D_{f}(x)= f^{'}(x)$ is invertible.(I don't know if it is true). I just prove that $||D_{f}(x)||\geq c$. It helps?
Thank you for any help.
 A: $f$ is surjective because $f(\mathbb R^n)$ is an open and closed subset of $\mathbb R^n$. You already mentioned that $f$ is an open map.  To see that $f(\mathbb R^n)$ is closed, let $(y_n)$ be a sequence in $f(\mathbb R^n)$ converging to $y$. Then because $\|f^{-1}(y_n)-f^{-1}(y_m)\|\leq c^{-1}\|y_n-y_m\|$, $(f^{-1}(y_n))$ is a Cauchy sequence, hence converges to some $x\in\mathbb R^n$, and $f(x) =f(\lim\limits_{n\to\infty}f^{-1}(y_n))=\lim\limits_{n\to\infty}y_n=y$, the second to last equality by continuity.  By connectedness, the only nonempty open and closed subset of $\mathbb R^n$ is $\mathbb R^n$.
You can show that $f'(x)$ is always invertible by contraposition.  Let $c>0$.  Suppose $f'(x)$ is not invertible for some $x$.  Let $v\in\mathbb R^n$ be a unit vector such that $f'(x)v=0$.  Then $\lim\limits_{h\to 0}\dfrac{\|f(x+hv)-f(x)\|}{|h|}=0$, so there exists $\delta>0$ such that $0<|h|<\delta$ implies $$\|f(x+hv)-f(x)\|<c|h|=c\|(x+hv)-x\|,$$ showing that the hypothesis doesn't hold.
Hence your $f$ does have $f'(x)$ invertible for all $x$, and the inverse function theorem therefore implies $f^{-1}$ is $C^1$.
