If $f :{\Bbb R}^2 \to {\Bbb R}^2$ is a function satisfying $\| f(x) - f(y) \| \geq 3\|x-y\|$ , then $f$ is an open mapping and generalizing it Let $f : {\Bbb R}^2 \to {\Bbb R}^2$ be a function such that $\| f(x) - f(y) \| \geq 3\|x-y\|$ ($\|\cdot\|$ denotes the usual metric in ${\Bbb R}^2$). Then I was trying to show that $f$ is an open mapping i.e. carries open sets onto open sets. 
My attempt:
[Let $A^\circ$ denote the interior of $A$ ($ \subseteq {\Bbb R}^2$).]
Since, in ${\Bbb R}^2$ every  open set is a countable union of open balls, the problem can be reduced to an open ball in ${\Bbb R}^2$ say, $B_r {(a)}$ (i.e. an open ball of radius $r$ centered at some $a \in {\Bbb R}^2$ ), then all I need to show is that, $$f({B_r {(a)}^\circ}) \subseteq {f(B_r {(a)})}^\circ$$  i.e. $$f({B_r {(a)}}) \subseteq {f(B_r {(a)})}^\circ.$$
Thanks in advance for help.
My second question is that how to generalize this  (and this makes my question different from the one identified by users **)i.e. **for which range of values of $n$ and $k$ , it is true that :
If $f :{\Bbb R}^n \to {\Bbb R}^n$ is a function satisfying $\| f(x) - f(y) \| \geq k\|x-y\|$ , then $f$ is an open mapping. ($\|\cdot\|$ denotes the usual metric in ${\Bbb R}^n$)
 A: Partial answer if $f$ is surjective.
$f$ is injective. $f(x)=f(y)$ implies that $\|f(x)-f(y)\|=0\geq 3\|x-y\|$.
Suppose it is not open, there exits an open subset $U$ such that $f(U)$ is not open, there exists a $y \in f(U), y=f(x), y_n$ which is not in $f(U)$ such that $lim_ny_n=y$. Write $y_n=f(x_n)$, $\|y_n-y_m\|\geq 3\|x_n-x_m\|$ this implies that $(x_n)$ is a Cauchy sequence and converges towards $x'$ such that $f(x')=y, x'=x$  since $f$ is injective, there exists $N$ such that for $n>N, x_n\in U$ since $U$ is open. contradicition.
A: This is not true as stated. Take 
$$
f(x_1,x_2) =\begin{cases} (3x_1, 3x_2),\quad & x_1<0  \\
(3x_1+1, 3x_2),\quad & x_1\ge0 
\end{cases}
$$
Then the inequality $\| f(x) - f(y) \| \geq 3\|x-y\|$ holds (the extra $+1$ only makes distances $f(x)-f(y)$ greater) but $f(\mathbb{R}^2)$ is not open, it is the set 
$$
\{(u, v)\in \mathbb{R}^2 : u\in (-\infty,0)\cup [1, \infty), \ v\in\mathbb{R}\}
$$

The statement is true if $f$ is also assumed continuous, but you are going to need some degree theory (winding number, in the planar case) to prove it.
