Solution
= Solution
Because $2|x|\leq1+x^2\leq Q$, we have $|f_x(x,y)|=2|x|/Q^2\leq1/Q\leq1$; similarly $|f_y|\leq1$. Apply the one-variable <mean value theorem> on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The <triangle inequality> gives
$$
\begin{aligned}
|f(x,y)-f(u,v)|&\leq|f(x,y)-f(u,y)|+|f(u,y)-f(u,v)|\\
&\leq\boxed{|x-u|+|y-v|}.
\end{aligned}
$$
Thus the function is globally one-Lipschitz for the $\ell^1$ metric on the plane.