Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be given by $f(\mathbf{x}) = \sqrt{1+\Vert\mathbf{x}\Vert^2}$, where $\Vert \cdot \Vert$ is the Euclidean norm.
I wish to show that $f \in C^{1,1}_1$, that is,
$$|f(\mathbf{x})-f(\mathbf{y})| \leq \Vert\mathbf{x}-\mathbf{y}\Vert$$ for $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$.
It is easy to show that
$$\left|\sqrt{1+\Vert\mathbf{x}\Vert^2} - \left(\sqrt{1+\Vert\mathbf{y}\Vert^2}\right)\right| \leq \left|\Vert{x}\Vert^2-\Vert{y}\Vert^2\right|,$$
where you simply multiply by a radical conjugate and remove the denominator to force the inclusion of the $\leq$ sign (since the denominator is always $\geq 1$).
However, I do not know how I would manipulate this to get to the final result. It is similar in form to the reverse triangle inequality, however that doesn't work for the square of norms (as far as I know). I am also unsure if my first few steps are even the correct steps to make in an attempt to answer the question.
Thank you.
What do you mean by $C^{1,1}_1$? It looks like you just want $C^{0,1}$.
You started out just fine but threw away some valuable information. Since $$\sqrt{1 + \|x\|^2} - \sqrt{1 + \|y\|^2} = \frac{\|y\|^2 - \|x\|^2}{\sqrt{1 + \|x\|^2} + \sqrt{1 + \|y\|^2}} = \frac{(\|y\|-\|x\|)(\|y\|+\|x\|)}{\sqrt{1 + \|x\|^2} + \sqrt{1 + \|y\|^2}}$$ you end up with $$ \left| \sqrt{1 + \|x\|^2} - \sqrt{1 + \|y\|^2} \right| = \frac{ {\|y\|+\|x\|}}{\sqrt{1 + \|x\|^2} + \sqrt{1 + \|y\|^2}} \bigg| \|x\| - \|y\| \bigg|.$$ The fraction is bounded by $1$. Now use the triangle inequality judiciously.