3. Solve the inequality $\sqrt{2 x+\frac{7}{x^{2}}}+\sqrt{2 x-\frac{7}{x^{2}}}<\frac{6}{x}$.

Answer. $x \in\left[\sqrt[3]{\frac{7}{2}} ; \sqrt[3]{\frac{373}{72}}\right)$.
The domain of definition (ODZ) is determined by the inequalities $2 x+\frac{7}{x^{2}} \geq 0, 2 x-\frac{7}{x^{2}} \geq 0$, solving which, we get $x \geq \sqrt[3]{\frac{7}{2}}$.

On the ODZ, both sides of the given inequality are positive, so we square them:

$$
\begin{aligned}
& 2 x+\frac{7}{x^{2}}+2 \sqrt{\left(2 x+\frac{7}{x^{2}}\right)\left(2 x-\frac{7}{x^{2}}\right)}+2 x-\frac{7}{x^{2}}<\frac{36}{x^{2}} \Leftrightarrow \sqrt{4 x^{2}-\frac{49}{x^{4}}}<\frac{18}{x^{2}}-2 x \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { 1 8 } { x ^ { 2 } } - 2 x \geq 0 , } \\
{ 4 x ^ { 2 } - \frac { 4 9 } { x ^ { 4 } } < \frac { 3 2 4 } { x ^ { 4 } } - \frac { 7 2 } { x } + 4 x ^ { 2 } }
\end{array} \stackrel { \text { ODZ } } { \Leftrightarrow } \left\{\begin{array}{l}
x^{3} \leq 9, \\
\frac{72}{x}<\frac{373}{x^{4}}
\end{array} \Leftrightarrow x^{3}<\frac{373}{72}\right.\right.
\end{aligned}
$$

Considering the ODZ, $x \in\left[\sqrt[3]{\frac{7}{2}} ; \sqrt{\frac{373}{72}}\right)$.