3. Solve the inequality $\sqrt{3 x-\frac{5}{x^{2}}}+\sqrt{3 x+\frac{5}{x^{2}}}<\frac{8}{x}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b655a94cefa8e5880135g-4.jpg?height=114&width=411&top_left_y=1254&top_left_x=177)
The domain of definition (ODZ) is determined by the inequalities $3 x+\frac{5}{x^{2}} \geq 0, 3 x-\frac{5}{x^{2}} \geq 0$, solving which, we get $x \geq 3 \sqrt{\frac{5}{3}}$.

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

$$
\begin{aligned}
3 x+ & \frac{5}{x^{2}}+2 \sqrt{\left(3 x+\frac{5}{x^{2}}\right)\left(3 x-\frac{5}{x^{2}}\right)}+3 x-\frac{5}{x^{2}}<\frac{64}{x^{2}} \Leftrightarrow \sqrt{9 x^{2}-\frac{25}{x^{4}}}<\frac{32}{x^{2}}-3 x \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \frac { 3 2 } { x ^ { 2 } } - 3 x \geq 0 , } \\
{ 9 x ^ { 2 } - \frac { 2 5 } { x ^ { 4 } } < \frac { 1 0 2 4 } { x ^ { 4 } } - \frac { 1 9 2 } { x } + 9 x ^ { 2 } }
\end{array} \Leftrightarrow \left\{\left\{\begin{array}{l}
\text { ODZ } 3 \\
\frac{192}{x}<\frac{1049}{x^{4}} \Leftrightarrow x^{3}<\frac{1049}{192} .
\end{array}\right.\right.\right.
\end{aligned}
$$

Considering the ODZ, $x \in\left[\sqrt[3]{\frac{5}{3}} ; \sqrt[3]{\frac{1049}{192}}\right)$.