$|f(x)-f(y)|\ge c|x-y|$, 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.

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*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.
