I am trying to show that
$|f(x)-f(y)|<|x-y|$,
for the function $f$ to be defined as $f:[0,+\infty)\mapsto [0,+\infty)$, $f(x)=(1+x^2)^{1/2}$, using the mean value theorem.
I have done this:
Since $f$ is differentiable on $[0,+\infty)$, then there is a point $x_0$, $x<x_0<y$, such that
$f(x)-f(y)=(x-y)f'(x_0)$,
by the mean value theorem. Hence,
$|f(x)-f(y)|=|x-y||f'(x_0)|=|x-y||x_0 (1+{x_0} ^2)^{-1/2}|\leq|x-y||x_0|\leq|x-y|M<|x-y|$
where M is a constant.
Can someone tell me if this is correct?