1. We want to solve the inequality $x < \sqrt{x^2 + 1}$.\n\n2. Notice that the square root expression $\sqrt{x^2 + 1}$ is always positive because $x^2 + 1 \geq 1 > 0$ for all real $x$.\n\n3. Square both sides to eliminate the square root, but be careful with inequalities. Since the right side is positive, squaring preserves the inequality direction for $x \geq 0$, but not necessarily for negative $x$. It is better to analyze by cases.\n\n4. Case 1: $x \geq 0$. Then squaring both sides is valid: $$x^2 < x^2 + 1.$$ Simplifying gives $$0 < 1,$$ which is always true. So for $x \geq 0$, the inequality holds.\n\n5. Case 2: $x < 0$. Here, $x$ is negative but $\sqrt{x^2 + 1}$ is positive, so: $$x < \sqrt{x^2+1} > 0,$$ which means a negative number is less than a positive number, which is always true.\n\n6. Therefore, the inequality $x < \sqrt{x^2+1}$ holds for all real numbers $x$.\n\nFinal answer:\n$$\boxed{\text{All real numbers } x \text{ satisfy } x < \sqrt{x^2+1}.}$$