<s>[INST] Solve inequality: $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ Rational method to solve $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ inequality?
I tried to lead fractions to a common denominator, but I think that this way is wrong, because  I had fourth-degree polynomial in the numerator.
 [/INST] I think I'm found nice solving method. If $x\neq 0$:
$$\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}  \equiv \frac{2}{x-2+\frac{5}{x}} + \frac{3}{x+2+\frac{5}{x}} \leq \frac{7}{8}$$
Let $t=x+\frac{5}{x}$.
$$\frac{2}{t-2} + \frac{3}{t+2} \leq \frac{7}{8}$$
$$\frac{(t-6)(7t+2)}{(t-2)(t+2)} \geq 0$$
$t \in (-\infty; -2) \cup [-\frac{2}{7}; 2) \cup [6; \infty)$
And finally solved the following inequalities:
$$x + \frac{5}{x} < -2$$
$$-\frac{2}{7}\leq x + \frac{5}{x} < 2$$
$$x + \frac{5}{x} \geq 6$$
I got $x \in  (-\infty; 1] \cup [5; \infty)$.
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