$|f(x)-f(y)| \geq \frac{|x-y|}{2}$ Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $|f(x)-f(y)| \geq \frac{|x-y|}{2}$ then prove $f$ is onto. I can prove it just using IVT, but looking for some short solution which is using some good argument.
 A: Here is a proof that uses the properties of homeomorphism in $\mathbb{R}$ and connectness. 
Let $U\subseteq \mathbb{R}$ be the image of $f$. 
Since $|f(x)-f(y)| \geq \frac{|x-y|}{2}$ it is imediate that $f$ is injective and $|f^{-1}(x)-f^{-1}(y)| \leq 2 |x-y|$. In particular, $f^{-1}$ is continuous.
So we have that $f$ is a homeomorphism from $\mathbb{R}$ onto $U$, so $U$ is an open interval in $\mathbb{R}$.
It is easy to show that 

if $A\subseteq \mathbb{R}$, then $\textrm{diam}(f(A))\geq \frac{1}{2}\textrm{diam}(A)$.

Now, remove $0$ from $\mathbb{R}$ then you have two disjoint connected components (intervals):
$$\mathbb{R}^-=\{x\in \mathbb{R} \,|\, x<0\}$$
and 
$$\mathbb{R}^+=\{x\in \mathbb{R} \,|\, x>0\}$$
Since $f$ is a homeomorphism, we have that removing $f(0)$ from $U$, we get two disjoint connected components (intervals): $U_1=f(\mathbb{R}^-)$ and $U_2=f(\mathbb{R}^+)$. But then we have: 
$$\textrm{diam}(U_1)=\textrm{diam}(f(\mathbb{R}^-))\geq \frac{1}{2}\textrm{diam}(\mathbb{R}^-)=+\infty$$
$$\textrm{diam}(U_2)=\textrm{diam}(f(\mathbb{R}^+))\geq \frac{1}{2}\textrm{diam}(\mathbb{R}^+)=+\infty$$
So $U$ is an interval such that removing one point from it, it is divided in two disjoint infinite intervals. It is easy to see that the only interval having such property is $\mathbb{R}$ itself. 
