$|f(x)-f(y)|\ge c|x-y|$, show that $Jf(x)\ne 0 \forall x\in \mathbb{R}^n$ and $f(\mathbb{R}^n)=\mathbb{R}^n$ $f$ is of class $C^{(1)}$ and there exist a number $c>0$ s.t. $|f(x)-f(y)|\ge c|x-y|$        $\forall x,y\in \mathbb{R}^n$. Show that $Jf(x)\ne 0 \forall x\in \mathbb{R}^n$ and $f(\mathbb{R}^n)=\mathbb{R}^n$,$Jf(x)$ is the Jacobian of $f(x)$.
Attempts:  I first let $Jf(x)=0$, then we can easily deduce that $|f(x)-f(y)-Df(x)(x-y)|\ge c|x-y|\implies |f(x)-f(y)|\ge (c+0)|x-y|$        $\forall x,y\in \mathbb{R}^n$ and $|f(x)-f(y)|<\epsilon|x-y|$ $\forall \epsilon>0$ whenever  $|x-y|<\delta$ (By differentiability) so we have $(c)<\epsilon$ for any $\epsilon>0$ and $|x-y|<\delta$ for some $\delta>0$ and hence $c=0$, contradiction. But it seem a little bit strange.
For the 2nd question, I have no idea how to prove that.
 A: *

*If the Jacobian determinant is zero at $x$, then $Df(x)$ has a nontrivial kernel. Let $v$ be a vector from the kernel. Show that $f(x+tv)-f(x)$ is small. 

*Inverse mapping theorem implies $f(\mathbb R^n)$ is open. The assumption $|f(x)-f(y)|\ge c|x-y|$ implies $f(\mathbb R^n)$ is closed.
A: I think a little elaboration of the above solution would be beneficial. For all $ x,v \in \mathbb{R}^n $ we have $ f(x+v )  = f(x) + Df(x)(v ) + o(\|v\|) $ and hence if $ Df(x)(v) = 0 $ then for $ h \in \mathbb{R} $ we have 
$$ c|h|\|v\| \leq \|f(x+hv)-f(x)\| \leq |h|\|Df(x)(v)\| +o(|h|) = o(|h|) $$ Thus on dividing by $ |h| $ and taking $ |h| \rightarrow 0 $ we have $ c\|v\|= 0 $ which gives $ v = 0 $, hence $ Jf(x) \neq 0 $ for all $ x \in \mathbb{R}^n $. Hence from inverse function theorem, there exists a neighborhood of $ f(x) $ where $f$ is invertible, so $f(\mathbb{R}^n)$ is open. Now for any sequence $ x_n $ if $ f(x_n) \rightarrow z $ then for any $ \epsilon > 0 $ and sufficiently large $n$ we have 
$$ \|x_n\| \leq \frac{1}{c}\|f(x_n)-f(0)\|\leq \frac{1}{c}(\|f(x_n)\|+\|f(0)\|) \leq \frac{1}{c}(\|z\|+\|f(0)\| +\epsilon) $$  So $ x_n $ is bounded sequence hence has a convergent subsequence $ x_{n_j} \rightarrow x $. Thus $ f(x_{n_j})\rightarrow f(x) = z $ from continuity. Hence $ f(\mathbb{R}^n) $ is closed. $\mathbb{R}^n $ being connected, we have $ f(\mathbb{R}^n) = \mathbb{R}^n $ 
