Prove that there exist $\mu>0$ and $\delta>0$ s.t. $|f(x)-f(y)|\geq \mu |x-y|$ Assume that $\Omega$ is an open set in $R^m$, $f \in C^1(\Omega,R^m)$, $a\in \Omega$. If $\det f'(a) \neq 0$, prove that there exist $\mu>0$ and $\delta>0$ s.t. for all $x,y \in B_{\delta}(a)$,  $|f(x)-f(y)|\geq \mu |x-y|$.

I can solve the situation when $m=1$. W.L.O.G. assume that $f'(a)>0$, then for $\epsilon = f'(a)/2$, there exists $\delta>0$ s.t. $\forall \xi \in B_{\delta}(a)$ we have $|f'(\xi)-f'(a)|<\epsilon$, hence $f'(\xi)>-\epsilon+f'(a)>0$. then by Lagrange's theorem, for all $x,y\in B_{\delta}(a)$, there exists $\xi =x+\theta(y-x)$ where $\theta \in [0,1]$ s.t. $|f(y)-f(x)|=|f'(\xi)(y-x)| \geq |-\epsilon+f'(a)|\cdot |y-x|$. Hence $\mu=-\epsilon+f'(a)$ and $\delta$ would meet our requirement.

But I have no idea how to deal with a higher dimensional space. Any help would be appreciated.
 A: In higher dimensions ( and even for general normed spaces) we have the inequality
$$\|g(x) - g(y) \|\le \sup_{t\in [x,y]} \|g'(t)\| \cdot \|x-y\|$$
Now apply this inequality for the function
$$g(x)\colon = f(x) - f'(a)x$$
We get
$$\tag{1} \|f(x)- f(y) - f'(a)(x-y)  \| \le \sup_{t\in [x,y]} \|f'(t)-f'(a)\|\cdot \|x-y\|$$
Note that for the linear map $L\colon = f'(a)^{-1}$ there exists $C>0$ such that we have
$\| L w \| \le C \|w\|$ for any $w$. Now apply for $w = f'(a)v$,  $\eta = \frac{1}{2C}$ and get $\|f'(a) v\|\ge 2 \eta \|v\|$
and so
$$\tag{2} \|f'(a) (x-y)\|\ge 2 \eta \|(x-y)\|$$
Now choose $\delta$ such that  $\|f'(t)-f'(a)\| \le \eta$ for all $t \in B_{\delta}(a)$. We therefore get for all $x$,$y$ in $B_{\delta}(a)$
$$\tag{3} \sup_{t\in [x,y]} \|f'(t)-f'(a)\|\cdot \|x-y\| \le \eta \|x-y\|$$
From $(1),(2),(3)$, using the triangle inequality, we get
$$\|f(x)-f(y)\|\ge \eta\cdot  \|x-y\|$$
for all $x$,$y$ in $B_{\delta}(a)$.
A: The inverse function theorem provides balls $B_r(a)$ and $B_s(f(a))$ such that $f^{-1}|_{B_s(f(a))}$ exists and is a continuously differentiable function, which satisfies $(f^{-1})'(f(x))=f'(x)^{-1}$ for each $x\in B_r(a).$ This implies in particular, that
$\tag1 |\det (f^{-1}){'}(f(a))|=\frac{1}{|\det f'(a)|}>0$
Now, $\overline {B_{s/2}(f(a))}$ is compact, so $|(f^{-1})'|$ has a finite maximum $M$ there, which is strictly positive by $(1).$
Finally, if we choose a ball $B_{\delta}(a)\subseteq f^{-1}(B_{s/2}(f(a))\subseteq B_r(a),$ then
$\tag2 x\neq y\in B_{\delta}(a)\Rightarrow f(x)\neq f(y)\in B_{s/2}(f(a))$
and the mean value inequality gives
$\tag3 |x-y|=|f^{-1}(f(x))-f^{-1}(f(y))|\le M|f(x)-f(y)|\Rightarrow \frac{|f(x)-f(y)|}{|x-y|}\ge \frac{1}{M}$
A: Here is another proof. Let $\phi: R^m \rightarrow R^m,x \mapsto x-A^{-1}f(x) $, where $A^{-1}=[f'(a)]^{-1}$. Then $\phi'(x)=I_m-A^{-1}f'(x)$, and  $\phi'(0)=0_m.$ Since the norm function is continuous, there exists $\delta >0$ s.t. if $\xi \in B_\delta (0)$ then $\big| \Vert \phi(\xi) \Vert-\Vert \phi'(0)\Vert \big|< 1/2,$ i.e, $ \Vert \phi(\xi) \Vert<1/2$. On the one hand, for $\forall x,y\in B_\delta (0)$ there exist $\xi \in (x,y)$ s.t.
$$
\big| \phi(y)-\phi(x) \big| \leq \Vert \phi'(\xi)\Vert \big| y-x \big| \leq1/2 \big| y-x \big|.
$$
On the other hand,
$$
\big| \phi(y)-\phi(x) \big|=\big| y-A^{-1}f(y)-x+A^{-1}f(x) \big| \geq \big|y-x \big| - \big| A^{-1}(f(x)-f(y)) \big| \\ \geq \big| y-x \big| -\Vert A^{-1}\Vert \big| (f(x)-f(y)) \big|.
$$
Hence,
$$
1/2 \big| y-x \big|\geq \big| y-x \big| -\Vert A^{-1}\Vert \big| (f(x)-f(y)) \big|, \\
\big| f(x)-f(y) \big| \geq \frac{1}{2\Vert A^{-1}\Vert}\big| y-x \big|.
$$
Thus we found $\delta$ and $\mu=\frac{1}{2\Vert A^{-1}\Vert} $.
