3. Solve the inequality $\frac{4 x^{4}+1}{4 \sqrt{2}} \leq x \sqrt{x^{4}-\frac{1}{4}}$.
Solution. Domain of definition: $x^{4}-\frac{1}{4} \geq 0, \quad x \in(-\infty ;-1 / \sqrt{2}] \cup[1 / \sqrt{2} ;+\infty)$

1) for $x \in(-\infty ;-1 / \sqrt{2}]$ the inequality has no solutions;
2) for $x \in[1 / \sqrt{2} ;+\infty)$ we square both sides of the inequality:

$$
\frac{16 x^{8}+8 x^{4}+1}{32} \leq x^{2}\left(x^{4}-\frac{1}{4}\right) \Leftrightarrow \quad 16 x^{4}+8+\frac{1}{x^{4}} \leq 8\left(4 x^{2}-\frac{1}{x^{2}}\right) \Leftrightarrow\left(4 x^{2}-\frac{1}{x^{2}}\right)^{2}+16 \leq 8\left(4 x^{2}-\frac{1}{x^{2}}\right)
$$

Let's make the substitution $t=4 x^{2}-\frac{1}{x^{2}}$. We arrive at the inequality $t^{2}-8 t+16 \leq 0 \Leftrightarrow(t-4)^{2} \leq 0 \Leftrightarrow t=4$.

After the reverse substitution, we have $4 x^{2}-\frac{1}{x^{2}}=4, \quad 4 x^{4}-4 x^{2}-1=0, \quad x^{2}=\frac{1+\sqrt{2}}{2}, \quad x=\sqrt{\frac{1+\sqrt{2}}{2}}$.

Answer: $\sqrt{\frac{1+\sqrt{2}}{2}}$.