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 $f(x)-f(y)=(x-y)f'(x_0)$, 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?
Answer
Hint:
First prove the general result: if $f:\mathbb{R}\to \mathbb{R}$ is a differentiable function and $|f'(x)| < M$ for all $x\in\mathbb{R}$ then for all $x,y\in\mathbb{R}$ the inequality $$|f(x)-f(y)| < M|x-y|$$ holds. The proof is very similar to what you have done in the question.
Next prove that if $f(x) = \sqrt{1+x^2}$ then $|f'(x)| < 1$. To do this consider $f'(x)^2 = \frac{x^2}{1+x^2}$.
Combinding the two results above gives the desired result.
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