I would ask for help regarding a problem with the parametric inequalities.
$\dfrac{(x+a)}{(a-1)}+\dfrac{(x-a)}{(a+1)}-\dfrac{x}{(a+1)}-\dfrac{2(x-1)}{(a-1)}\ge 0\ \text{for}\ a<-1$
Since $a<-1$, both $(a-1)$ and $(a+1)$ will be negative. The lowest common denominator will be $(a-1)(a+1).$ I need to change the sense of the inequality.
$(x+a)(a+1)+(x-a)(a-1)-x(a-1)-2(x-1)(a+1)\le 0$
$-ax+4a-x+2\le 0$
$-ax-x\le -4a-2$
I decide to change the sign to both sides of the inequality
$ax+x\ge 4a+2$
$x(a+1)\ge 2(2a+1)$
Now divide by $(a+1)$ that being negative requires me to change the sign of the inequality
$x\le \dfrac{2(2a+1)}{(a+1)}$
But this is not the solution, because it is $x\ge \dfrac{2(2a+1)}{(a+1)}$. I can't understand what I'm doing wrong. I would be grateful if you could help. Sorry for my bad English.
$\dfrac{x+a}{a-1}+\dfrac{x-a}{a+1}-\dfrac{x}{a+1}-\dfrac{2(x-1)}{a-1}\ge 0$ is equivalent to
$\dfrac{-a}{a+1}+\dfrac{x+a-2x+2}{a-1}\ge 0\iff \dfrac{-x+a+2}{a-1}\ge \dfrac{a}{a+1}$. Then we have: $$-x+a+2\le \dfrac{a(a-1)}{a+1} \text{(since (a-1)<0, we flip the sign)}$$ $$\iff-x\le \dfrac{-(a+2)(a+1)+a(a-1)}{a+1}\iff -x\le \dfrac{-4a-2}{a+1}\iff -x\le \dfrac{-2(2a+1)}{a+1}\iff x\ge \dfrac{2(2a+1)}{a+1}$$