Show that $f$ is $C^{1}$, $Df(0) = I$ implies that $\lVert f(x) - f(y) \rVert \geq \frac{1}{2}\lVert x-y\rVert$ Show that given $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is $C^{1}$, $Df(0) = I$, then $\lVert f(x) - f(y) \rVert \geq \frac{1}{2}\lVert x-y\rVert$ for $x,y$ in a sufficiently small $\delta$ neighborhood about $0$.
How can I show this explicitly?
My intuition:
As someone put it, the moral of all differential calculus is that where the derivative is defined for a function, the function behaves as the linearization given by the derivative. This particular case is no different, I believe; given that all partials $\frac{\delta f_{i}}{\delta x_{j}}$ are $1$, locally at $0$, $a = [x,y]^{T}$ maps (approximately) to $b = [x+y, x+y]^{T}$. Thus, the distance between $a$ and $b$ is at least $$(y^{2} + x^{2})^{1/2} = \lVert a\rVert$$. Working with another set of points and a similar equation as above should give something meaningful.
 A: Note 
\begin{eqnarray}
f(y)&=&f(x)+Df(x)(x-y)+R_2(\|y-x\|)\\
&=&f(x)+Df(0)(x-y)+R_1(\|x\|)(x-y)+R_2(\|y-x\|).
\end{eqnarray}
where
$$ \lim_{x\to 0}R_1(\|x\|)=0,\lim_{y\to x}\frac{R_2(\|y-x\|)}{\|y-x\|}=0.$$
So there is small $\delta>0$ such that, for $x,y\in B_\delta(0)$,
$$\|R_1(\|x\|)\|<\frac14, \|R_1(\|x\|)\|<\frac14\|y-x\|. $$
Thus for $x,y\in B_\delta(0)$,
$$ \|f(y)-f(x)\|\ge\|y-x\|-\|R_1(\|x\|)\|\|y-x\|-\|R_2(\|y-x\|)\ge\frac12\|y-x\|.  $$
A: Credit to xpaul for solving this. I have accepted his answer. I think his proof needs a bit of polishing. I've included it below.
Note 
\begin{eqnarray}
f(y)&=&f(x)+Df(x)(x-y)+R(\|y-x\|)\\
&=&f(x)+Df(0)(x-y)+O(\|x\|)(x-y)+R_2(\|y-x\|).
\end{eqnarray}
where
$$ \lim_{x\to 0}O(\|x\|)=0,\quad lim_{y\to x}\frac{R_2(\|y-x\|)}{\|y-x\|}=0.$$
So there is small $\delta>0$ such that, for $x,y\in B_\delta(0)$,
$$\|R_1(\|x\|)\|<\frac14, \|R_1(\|x-y\|)\|<\frac14\|y-x\|. $$
Thus for $x,y\in B_\delta(0)$,
$$ \|f(y)-f(x)\|\ge\|y-x\|-\|R_1(\|x\|)\|\|y-x\|-\|R_2(\|y-x\|)\ge\frac12\|y-x\|.  $$
