Let $I:=[0,1]\times[-1,1]$, $\Vert.\Vert_\infty$ be a maximum norm and for $x=(x_1,x_2)$ $$\Phi(x)=\frac{1}{6}\begin{pmatrix} x_1e^{-x_2^2}+x_1x_2+3 \\ \log(1+x_1^2+x_2^2)-1 \end{pmatrix}$$ Show that $\Vert\Phi(x)-\Phi(y) \Vert_\infty \leq \frac{5}{6}\Vert x-y \Vert_\infty$ for $x,y\in I$.
My take: Let $x=(a,b)\in I$, $y=(c,d) \in I$ \begin{align} 6\Vert\Phi(x)-\Phi(y) \Vert_\infty&=\Vert \begin{pmatrix} ae^{-b^2}+ab -ce^{-d^2}-cd \\ \log(1+a^2+b^2) - \log(1+c^2+d^2) \end{pmatrix}\Vert_\infty \\ &= \max(|a(e^{-b^2}+b) -c(e^{-d^2}+d)|,|\log(\frac{1+a^2+b^2}{1+c^2+d^2}|)) \\ &\leq \max(2|a -c|,\log(3))\\ \end{align} I am not sure how to proceed from here. Is $\log(3)$ estimate too big?
One thing that you can do is considering the function $f(t): [0,2] \to \mathbb{R}$ defined by $f(t)=\log(1+t)$. By the mean value theorem we have that $$|\log(1+t_1) - \log(1+t_2)| \le |t_1 - t_2|.$$ Applied to our case we get $$ \begin{split}|\log(1+a^2 + b^2) - \log(1+c^2 + d^2)| & \le |a^2 + b^2 - c^2 - d^2 | \\ & \le 2 ( |a-c| + |b-d|) \\ & = 2 \| (x - y)\|_1 \\ & \le 4 \| (x - y)\|_\infty. \end{split} $$