Example 16. Solve the inequality

$$
-|y|+x-\sqrt{x^{2}+y^{2}-1} \geqslant 1
$$
Solution. The domain of the inequality (13) is determined by the condition $x^{2}+y^{2}-1 \geqslant 0$. Rewrite inequality (13) as

$$
x-|y| \geqslant 1+\sqrt{x^{2}+y^{2}-1}
$$

From this, it follows that $x-|y| \geqslant 0$. Under this condition, both sides of the obtained inequality are non-negative on its domain. Therefore, after squaring both sides of inequality (13) on its domain, we obtain an equivalent system

$$
\left\{\begin{array}{l}
-x|y| \geqslant \sqrt{x^{2}+y^{2}-1} \\
x \geqslant|y|
\end{array}\right.
$$

Since $x \geqslant|y| \geqslant 0$, the left side of the first inequality in this system is non-positive, and the right side is non-negative. Therefore, the system is satisfied only when both sides are equal to zero, i.e., inequality (13) is equivalent to the system

$$
\left\{\begin{array}{l}
x|y|=0 \\
x^{2}+y^{2}-1=0 \\
x \geqslant|y|
\end{array}\right.
$$

The first equation of the last system means that either $x=0$ or $y=0$.

If $x=0$, then from the inequality $x \geqslant|y|$ we find that $y=0$. But the pair $(0 ; 0)$ does not belong to the domain of the original inequality.

If $y=0$, then from the second equation we get (considering $x \geqslant 0$) $x=1$. The pair $(1 ; 0)$ satisfies inequality (13) and is its only solution.