For what values of $x$ does the following inequality hold?

$$
x^{2}-2 \sqrt{x^{2}+1}-2 \geq 0 .
$$
Since $x^{2}+1>0$, the left side is defined for all real $x$. The left side can be factored as:

$$
\left(\sqrt{x^{2}+1}-1\right)^{2}-2^{2}=\left(\sqrt{x^{2}+1}-3\right)\left(\sqrt{x^{2}+1}+1\right)
$$

The product is positive if either both factors are positive or both are negative. Since $\sqrt{x^{2}+1} \geq 0$, the second factor can never be negative, so only the inequality $\sqrt{x^{2}+1} \geq 3$ needs to be satisfied. Since both sides are positive, we can square them:

$$
x^{2}+1 \geq 9, \quad \text { hence } \quad x \geq \sqrt{8}, \quad \text { or } \quad x \leq-\sqrt{8}
$$

Based on the work of Éva Végh (Szekszárd, Garay J. Gymn., IV. grade).

Remarks. Squaring an inequality is not an equivalent transformation, e.g., $-3>-5$, but $(-3)^{2}<(-5)^{2}$.

Many reasoned as follows: $x^{2}-2 \geq 2 \sqrt{x^{2}+1}$, since the right side is non-negative, squaring it leads to an equivalent equation. This is only true if $x^{2}-2 \geq 0$. Consider, for example, the simple case: $x+1 \geq 1$. (Its solution is clearly $x \geq 0$.) Now square and rearrange; we get $x^{2}+2 x \geq 0$, and its solutions are $x \geq 0$ and $x \leq-2$. The latter is a false result.

Many forgot to determine the domain of definition, which is also necessary due to the square root.

Verifying the inequality with numerical examples is a meaningless task, since trying a finite number of values only allows us to determine whether the inequality holds for those values, and we can say nothing about all other numbers.