Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be continuously differentiable. If there exists a real number $a>0$, such that $\forall x,y\in \mathbb{R}^n$, $d\big(f(x),f(y)\big)\geq a\cdot d(x,y)$, please show that $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is a $C^1$-diffeomorphism.
It's clear that $f$ is injective, and the differential $df(x)$ is invertible for any $x\in \mathbb{R}$, since $f(x+v)=f(x)+df(x)(v)+o(v)$, as $v\rightarrow 0$. Then by the inverse function theorem, we have $f$ is a locally diffeomorphism. If $f$ is surjective, then we can show that $f$ is a $C^1$-diffeomorphism.
How can I show that $f$ is surjective?
I hope that you are familiar with some metric topology. The strategy is to prove that the image of $f$ is both open and closed as a subset of $\mathbb{R}^n$. Since $\mathbb{R}^n$ is connected, it is the only non-empty subset of itself having these two properties.
We first show that $U = f(\mathbb{R}^n)$ is closed in $\mathbb{R}^n$. Let a sequence $y_n\in U$ converge to some $y \in \mathbb{R}^n$. Then there exist $x_n$ such that $y_n = f(x_n)$. By the assumption we have $d(x_n,x_m) \leq a\cdot d(y_n,y_m)$, but since $y_n$ converges in $\mathbb{R}^n$ it is a Cauchy sequence, and thus $x_n$ is Cauchy as well and converges to some $x\in \mathbb{R}^n$. It follows that $y = f(x)\in U$.
The openness follows from the inverse function theorem. It states that the function is invertible on a neighborhood (that is, an open subset of $\mathbb{R}^n$) of each point $f(y)$, which implicitly says that the neighborhood is contained in the image (cf. the proof of Theorem 2 here).