Solve the following inequality over the set of real numbers:

$$
x^{2}-3 \sqrt{x^{2}+3} \leq 1
$$
Solution. The expression under the square root, $x^{2}+3$, is positive. Let's introduce the new variable $\sqrt{x^{2}+3}=a$. Then $a \geq \sqrt{3}$ and we arrive at the inequality

$$
a^{2}-3-3 a \leq 1
$$

Factoring, we get $(a-4)(a+1) \leq 0$. The inequality holds when $-1 \leq a \leq 4$. The condition $\sqrt{x^{2}+3} \geq \sqrt{3}>-1$ is always satisfied. If $a \leq 4$, then $\sqrt{x^{2}+3} \leq 4$. Since neither side is negative, we can square both sides:

$$
x^{2}+3 \leq 16 \quad \text { and hence } \quad-\sqrt{13} \leq x \leq \sqrt{13}
$$

This is the solution to the inequality.