Task B-2.1. In the set $\mathbb{R}$, solve the inequality $\left(\frac{x}{x+1}\right)^{2}+\left(\frac{x}{x-1}\right)^{2} \geq 6$.
## Solution.

If we multiply the inequality $\left(\frac{x}{x+1}\right)^{2}+\left(\frac{x}{x-1}\right)^{2} \geq 6 \text{ with } (x+1)^{2}(x-1)^{2}$, under the condition $x \neq 1, x \neq-1$, we get:

$$
\begin{aligned}
& x^{2}(x-1)^{2}+x^{2}(x+1)^{2} \geq 6(x+1)^{2}(x-1)^{2} \\
& x^{2}\left[(x-1)^{2}+(x+1)^{2}\right] \geq 6\left(x^{2}-1\right)^{2} \\
& x^{2}\left[2 x^{2}+2\right] \geq 6\left(x^{2}-1\right)^{2} \\
& x^{2}\left[x^{2}+1\right]-3\left(x^{2}-1\right)^{2} \geq 0 \\
& -2 x^{4}+7 x^{2}-3 \geq 0 \text { or } 2 x^{4}-7 x^{2}+3 \leq 0
\end{aligned}
$$

By solving the biquadratic equation $2 x^{4}-7 x^{2}+3=0$, we get

$$
x^{2}=3 \text { and } x^{2}=\frac{1}{2}, \text { i.e., } x_{1,2}= \pm \sqrt{3} \text { and } x_{3,4}= \pm \frac{\sqrt{2}}{2}
$$

We write

$$
2\left(x^{2}-3\right)\left(x^{2}-\frac{1}{2}\right) \leq 0 \text { or } 2(x+\sqrt{3})(x-\sqrt{3})\left(x+\frac{\sqrt{2}}{2}\right)\left(x-\frac{\sqrt{2}}{2}\right) \leq 0
$$

Using a graph (or a table), the solution to the inequality is

$$
x \in\left[-\sqrt{3},-\frac{\sqrt{2}}{2}\right] \cup\left[\frac{\sqrt{2}}{2}, \sqrt{3}\right] \backslash\{-1,1\}
$$