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?

$$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{x-ax+a+1}{x^2-1}\geq 0$$
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 $(-\infty, -1)\cup \left(1,\frac{a+1}{a-1}\right]$.
