<s>[INST] 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?
 [/INST] $$\frac{(x+1-ax+1)}{x^2-1}\ge0$$
$$\frac{x(1-a)+2}{x^2-1}\ge0$$
Since $a>1$
$$x\ge\frac{-2}{a-1}$$
$$x\le\frac{2}{a-1}\tag{x not equal to +,-1}$$ 
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