<s>[INST] Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$ Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$
To show a function is monotonically increasing, I started by saying that:

A function $f$ is monotonically increasing in an interval $(a, b)$ if for all $x, y \in (a, b)$, with $x$ < $y$, we have that $f(x) \leq f(y)$.

So I started by tryin to show:

Let $x$, $y$ be arbitrary fixed numbers in $(-1, 1)$, with $x < y$, we want to show that $f(x) \leq f(y)$:
  $$f(x) = \frac{1 - 2x}{x^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} = f(y)$$
  We multiply the left side by $\frac{y^2 - 1}{y^2 - 1}$, which is positive because $y^2$ is always less than $1$, so we don't need to change the sign:
  $$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1}$$
  We know multiply by $\frac{x^2 - 1}{x^2 - 1}$ the right side:
  $$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} \cdot \frac{x^2 - 1}{x^2 - 1}$$
  So we can simplify:
  $$(1 - 2x)\cdot (y^2 - 1) \leq (1 - 2y) \cdot (x^2 - 1)$$

Now, how do we proceed? Is this correct?
 [/INST] A differentiable function in some interval is monotonically aascending if its derivative is positive:
$$f'(x)=\frac{-2(x^2-1)-2x(1-2x)}{(x^2-1)^2}=\frac{-2x^2+2-2x+4x^2}{(x^2-1)^2}=$$
$$=2\frac{x^2-x+1}{(x^2-1)^2}$$
Now just prove the above indeed is positive for all $\;x\in\Bbb R\setminus\{-1,1\}\;$ , in particular in $\;(-1,1)\;$ .
Another way: Suppose $\;x<y\;$ , then ( observe that $\;t^2-1<0\;$ for $\;t\in(-1,1)\;$ )
$$f(x)<f(y)\iff\frac{1-2x}{x^2-1}<\frac{1-2y}{y^2-1}\iff y^2-1-2xy^2+2x>x^2-1-2x^2y+2y$$
$$\iff(y-x)\left[(y+x)-2xy-2\right]>0$$
Since $\;y-x>0\;$ , it is enough to show the other factor is positive: this part is for you.
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