3. $\left\{x \left\lvert\, x>\frac{\sqrt[3]{10}}{2}\right.\right\}$

Analysis: From $x+\frac{1}{x^{2}} \geq 0, x-\frac{1}{x^{2}} \geq 0$ we get $x \geq 1$, the original inequality is equivalent to $\sqrt{x^{3}+1}-\sqrt{x^{3}-1}\frac{1}{2} \Rightarrow x>\frac{\sqrt[3]{10}}{2}$, thus the solution set of the inequality is $\left\{x \left\lvert\, x>\frac{\sqrt[3]{10}}{2}\right.\right\}$
3. The solution set of the inequality $\sqrt{x+\frac{1}{x^{2}}}-\sqrt{x-\frac{1}{x^{2}}}<\frac{1}{x}$ is