9・13 Solve the inequality
$$\sqrt{\frac{\pi}{4}-\operatorname{arctg} \frac{|x|+|y|}{\pi}}+\operatorname{tg}^{2} x+1 \leqslant \sqrt{2}|\operatorname{tg} x|(\sin x+\cos x)$$
[Solution] Let $x, y$ satisfy the inequality. From $\sin x + \cos x \leqslant \sqrt{2}$, we have
$$\begin{aligned}
\tan^2 x + 1 & \leqslant \sqrt{\frac{\pi}{4} - \arctan \frac{|x| + |y|}{\pi}} + \tan^2 x + 1 \\
& \leqslant \sqrt{2}|\tan x|(\sin x + \cos x) \\
& \leqslant 2|\tan x|
\end{aligned}$$

Since $\tan^2 x + 1 \geqslant 2|\tan x|$, we have
$$\begin{aligned}
\tan^2 x + 1 & = \sqrt{\frac{\pi}{4} - \arctan \frac{|x| + |y|}{\pi}} + \tan^2 x + 1 \\
& = \sqrt{2}|\tan x|(\sin x + \cos x) \\
& = 2|\tan x|
\end{aligned}$$

From this, we can deduce
$$\begin{array}{l}
\frac{\pi}{4} - \arctan \frac{|x| + |y|}{\pi} = 0 \\
|\tan x| = 1, \sin x + \cos x = \sqrt{2}
\end{array}$$

Thus, $x = \frac{\pi}{4} + 2n\pi, n \in \mathbb{Z}, |x| + |y| = \pi$, which implies $n = 0$, i.e., $x = \frac{\pi}{4}$, $y = \pm \frac{3}{4} \pi$.

Conversely, when $x = \frac{\pi}{4}, y = \pm \frac{3}{4} \pi$, the inequality clearly holds. Therefore, the solutions are $(x, y) \in \left\{\left(\frac{\pi}{4}, \frac{3\pi}{4}\right), \left(\frac{\pi}{4}, -\frac{3\pi}{4}\right)\right\}$.