<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] 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?
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