Counter-example to contraction principle

One of the conditions of contraction principle is $d(fx,fy)<cd(x,y)$ for all $x,y$ where $c<1$. Now I am finding an example that when the condition is replaced by $d(fx,fy)<d(x,y)$, the theorem fails, i.e. we can not find a fixed point.

I am sure this is true.But I can not find an example.

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@derpy The typography was corrupted because of bad use of < characters outside math mode. See edited question. – Jeppe Stig Nielsen Apr 23 '14 at 14:16

The hypotheses to Banach-Caccioppoli fixed point theorem are the following: suppose $(X,d)$ is a complete metric space (that is, all $d$-Cauchy sequences converge in $X$), $X \ne \varnothing$, and let a function $f : X \to X$ be provided such that $$d(f(x),f(y))\le Cd(x,y) \qquad (*)$$ for all $x,y \in X$, where $C \in [0,1)$ is a constant which doesn't depend on $(x,y)$ (such $f$ is called a contraction of $X$); then $f$ admits a fixed point, that is, there is an $x_0 \in X$ with $f(x_0) = x_0$.
The hypothesis $(*)$ may not in general be replaced by the weaker hypothesis $$d(f(x),f(y)) < d(x,y) \qquad (**)$$ for all $x,y \in X$, as shown in the following example.
Let $X = [1,+\infty)$ and $d(x,y) = \left|x-y\right|$; notice that $(X,d)$ is a closed subspace of the Banach space $(\mathbb R,d_\mathbb R)$, hence complete. Let $f : X \to X$ be given by $$f(x) = x + \frac{1}{x}.$$
Notice that $$\left|f(x) - f(y)\right| = \left|x - y - \frac{x - y}{xy}\right| = \left|(x - y)\left(1 - \frac{1}{xy}\right)\right| < \left|x - y\right|$$ whenever $x,y \in X$. Still, you can verify that $f$ doesn't fix any point in $X$. The failure can be appointed to $f$ fixing the "point at infinity"; actually, hypothesis $(**)$ is sufficient if $X$ happens to be compact.