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.