1. **State the problem:** We need to analyze the inequality $$\frac{5}{x^2 + 1} \leq 1$$ and find the values of $x$ that satisfy it. 2. **Rewrite the inequality:** Since $x^2 + 1 > 0$ for all real $x$, we can multiply both sides by $x^2 + 1$ without changing the inequality direction: $$\frac{5}{x^2 + 1} \leq 1 \implies 5 \leq x^2 + 1$$ 3. **Simplify the inequality:** $$5 \leq x^2 + 1 \implies 5 - 1 \leq x^2 \implies 4 \leq x^2$$ 4. **Solve for $x$:** $$x^2 \geq 4 \implies x \leq -2 \text{ or } x \geq 2$$ 5. **Interpretation:** The solution to the inequality is all $x$ such that $x \leq -2$ or $x \geq 2$. **Final answer:** $$x \leq -2, \quad x \geq 2$$