$\ln(1+x_1^2 + x_2^2)$ is Lipschitz in $\mathbb{R}^2$?

Let $x := (x_1,x_2) \in \mathbb{R}^2$, is the mapping $x \mapsto \ln(1+x_1^2+x_2^2)$ is Lipschitz w.r.t to the Euclidean $2$-norm?

My attempt: Let $x,y \in \mathbb{R}^2$, then we have: $$|\ln(1+x_1^2+x_2^2) - \ln(1+y_1^2+y_2^2)| \le \Big|\frac{x_1^2+x_2^2-y_1^2-y_2^2}{1+y_1^2+y_2^2}\Big|$$ by the inequality that $\ln(1+x) \le x$, but how the proceed next step? I appreciate your help, thanks

• Immediately it would seem simpler to prove that the partial derivatives exist everywhere and are globally bounded. – Henning Makholm Apr 2 '18 at 23:18

Write $f(x_{1},x_{2}) = \ln(1+x_{1}^{2}+x_{2}^{2})$. Then the partial derivatives of $f$ are $$\frac{\partial f}{\partial x_{i}}= \frac{2x_{i}}{1+x_{1}^{2}+x_{2}^{2}}, \quad i=1,2$$
For $x_{i} \leq 2$ we have $$\frac{\partial f}{\partial x_{i}} \leq \frac{4}{1+x_{1}^{2}+x_{2}^{2}}\leq 4$$
and for $x_{i}>2$ we have $2x_{i}<x_{i}^{2}$ so that $$\frac{\partial f}{\partial x_{i}} < \frac{x_{i}^{2}}{1+x_{1}^{2}+x_{2}^{2}} < 1$$
Because the partial derivatives are defined everywhere and are bounded, $f$ is Lipschitz.