Solve inequality: $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$

Question

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.

Answer 1

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 $x \in (-\infty; 1] \cup [5; \infty)$.

Answer 2

HINT:

As $\displaystyle x^2\pm2x+5=(x\pm1)^2+4\ge4>0$ for real $x$

we can safely multiply out either sides by $(x^2+2x+5)(x^2-2x+5)$

Then find the roots of the Quartic Equation