1. Zadatak

Pomnožimo li nejednadžbu

$$\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2\ge6$$

s $(x+1)^2(x-1)^2$, uz uvjet $x\ne1$, $x\ne-1$ redom slijedi:

$$x^2(x-1)^2+x^2(x+1)^2\ge6(x+1)^2(x-1)^2$$ $$x^2\big[(x-1)^2+(x+1)^2\big]\ge6(x^2-1)^2$$ $$x^2\big[2x^2+2\big]\ge6(x^2-1)^2$$ $$x^2\big[x^2+1\big]-3(x^2-1)^2\ge0$$ $$-2x^4+7x^2-3\ge0\quad\text{ili}\quad 2x^4-7x^2+3\le0.$$

Rješavanjem bikvadratne jednadžbe $2x^4-7x^2+3=0$ dobivamo

$$x^2=3\ \text{i}\ x^2=\frac12, \qquad\text{odnosno}\qquad x_{1,2}=\pm\sqrt3\ \text{i}\ x_{3,4}=\pm\frac{\sqrt2}{2}.$$

Pišemo

$$2\left(x^2-3\right)\left(x^2-\frac12\right)\le0$$

ili

$$2(x+\sqrt3)(x-\sqrt3)\left(x+\frac{\sqrt2}{2}\right)\left(x-\frac{\sqrt2}{2}\right)\le0.$$

Pomoću grafa (ili tablice) slijedi rješenje nejednadžbe

$$x\in\left[-\sqrt3,-\frac{\sqrt2}{2}\right]\cup\left[\frac{\sqrt2}{2},\sqrt3\right]\setminus\{-1,1\}.$$