Example 5 Solve the inequality $\sqrt{\frac{\pi}{4}-\arctan \frac{|x|+|y|}{\pi}}+\tan ^{2} x+1 \leqslant \sqrt{2}|\tan x|(\sin x+\cos x)$.


Since $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \leqslant \sqrt{2}$, and $\sqrt{\frac{\pi}{4}-\arctan \frac{|x|+|y|}{\pi}} \geqslant 0$, we have, from the given problem, $\tan ^{2} x+1 \leqslant \sqrt{\frac{\pi}{4}-\arctan \frac{|x|+|y|}{\pi}}+\tan ^{2} x+1 \leqslant \sqrt{2}|\tan x|(\sin x+\cos x) \leqslant$ $2|\tan x|$.
However, $(|\tan x|-1)^{2} \geqslant 0$, which means $\tan ^{2} x+1 \geqslant 2|\tan x|$, so the inequalities above must all be equalities, i.e., $\sqrt{\frac{\pi}{4}-\arctan \frac{|x|+|y|}{\pi}}=0, \sqrt{2} \sin \left(x+\frac{\pi}{4}\right)=\sqrt{2},|\tan x|=1$.
Thus, we get $|x|+|y|=\pi$, and $x=2 k \pi+\frac{\pi}{4}, k \in \mathbf{Z}$.
Since $|x| \leqslant \pi$, i.e., $-\pi \leqslant x \leqslant \pi$, we have $k=0$, so $x=\frac{\pi}{4}, y= \pm \frac{3}{4} \pi$. Therefore, the solutions to the original inequality are $(x, y)=\left(\frac{\pi}{4}, \frac{3}{4} \pi\right)$ or $\left(\frac{\pi}{4},-\frac{3}{4} \pi\right)$.