Solve $\frac{1}{x-1}\geq\frac{a}{x+1}$, provided $a>1$ Moving everything to the left,
$$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{ax-x-a-1}{x^2-1}\leq 0.$$
Finding roots to nominator and denominator:
$$\begin{array}{lcl}
ax-x-a-1 & = & 0 \Leftrightarrow \frac{a+1}{a-1}\\
x^2-1 & = & 0 \Leftrightarrow x_1 =1  \ \text{och} \ x_2=-1.\\ 
\end{array}$$
And that's it. How to proceed?
 A: $$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{x-ax+a+1}{x^2-1}\geq 0$$
Last inequality is the same that
$$\frac{(a-1)\left(-x+\frac{a+1}{a-1}\right)}{(x+1)(x-1)}\ge 0\quad\iff\quad \frac{-x+\frac{a+1}{a-1}}{(x+1)(x-1)}\ge 0$$
Now, we can test for $x$ in each interval $(-\infty,-1)$, $(-1,1)$, $\left(1,\frac{a+1}{a-1}\right]$ and $\left[\frac{a+1}{a-1},\infty\right)$. Which give us that the solution set is $(-\infty, -1)\cup \left(1,\frac{a+1}{a-1}\right]$.
A: Just analise the signal of each function: $f(x)=ax-x-a-1=x(a-1)-(a+1)$ and $g(x)=x^2-1$. 
Looking to the roots $-1,\frac{a+1}{a-1}, 1$ we have to know what is the relation between them.
Once $a>1$ then $\frac{a+1}{a-1}>0$ and then $\frac{a+1}{a-1}>-1$. It is also easy to see that $\frac{a+1}{a-1}>1$. So the relation is 
$$-1<1<\frac{a+1}{a-1}$$
Now we know that 
$$f(x)\ge0 \text{ if } x\ge \frac{a+1}{a-1}$$
$$f(x)\le 0\text{ if } x\le \frac{a+1}{a-1}$$
also,
$$g(x)<0 \text{ if } -1<x<1$$
$$g(x)>0 \text{ if } x<-1 \text{ or } x>1$$
so,
$1)$ For $x<-1$ we have $f(x)<0$ and $g(x)>0$ so $\frac{f(x)}{g(x)}<0$;
$2)$ For $-1<x<1$ we have $f(x)<0$ and $g(x)<0$ so $\frac{f(x)}{g(x)}>0$;
$3)$ For $1<x\le \frac{a+1}{a-1}$ we have $f(x)\le0$ and $g(x)>0$ so $\frac{f(x)}{g(x)}\le0$;
$4)$ For $x\ge \frac{a+1}{a-1}$ we have $f(x)\ge0$ and $g(x)>0$ so $\frac{f(x)}{g(x)}\ge0$;
So, the solution comes from $(1)$ and $(3)$. Just see that $\frac{a+1}{a-1}=1+\frac{2}{a-1}$.
A: $$\frac{(x+1-ax+1)}{x^2-1}\ge0$$
$$\frac{x(1-a)+2}{x^2-1}\ge0$$
Since $a>1$
$$x\ge\frac{-2}{a-1}$$
$$x\le\frac{2}{a-1}\tag{x not equal to +,-1}$$ 
A: you have to solve $$\frac{x+1-a(x-1)}{x^2-1}\geq 0$$
the first case is given by
$$x+1-ax+a\geq 0$$ and $$x^2>1$$
this gives
$$x(1-a)\geq -(a+1)$$ and $$|x|>1$$
since $$a>1$$ we get
$$x\le \frac{a+1}{1-a}$$ and if $$x\geq 0$$ we get
the solution set
$$1<x\le \frac{a+1}{a-1}$$
can you proceed?
