For $0 \leq a,b \leq 1$, why does $aFor $0 \leq a,b \leq 1$, why does $a<b$ imply $\frac a{1-a }< \frac b{1-b}$?
Do we need to compute derivatives to validate this inequality? Or talk about increasing/decreasing functions?
 A: $$a<b \\\Rightarrow a-ab<b-ab\\\Rightarrow a(1-b)<b(1-a)\\\Rightarrow\frac{a}{1-a}<\frac{b}{1-b}$$
All calculations are valid because $0\leq a, b \leq 1$
A: Observe
\begin{align}
a-ab<b-ab.
\end{align}
A: Note that $\frac 1{\frac x{1-x}} = \frac 1x - 1$.
Clearly $\frac 1a > \frac 1b$. Now subtract $1$ from both sides, take the reciprocal (and flip the inequality).
A: If $0\le a\lt b\le1$, then the numerators and denominators of $a/(1-a)$ and $b/(1-b)$ are all nonnegative (and all positive if $0\lt a\lt b\lt1$). Now the numerator of $b/(1-b)$ is greater than the numerator of $a/(1-a)$, and the denominator of $b/(1-b)$ is smaller than the denominator of $a/(1-a)$. Therefore the fraction $b/(1-b)$ is larger than the fraction $a/(1-a)$. If you like,
$${a\over1-a}\lt{a\over1-b}\lt{b\over1-b}$$
or, if you prefer,
$${a\over1-a}\lt{b\over1-a}\lt{b\over1-b}$$
(Note, we are presumably meant to accept $b/(1-b)=\infty$ if $b=1$.)
A: Rearrange the inequality as follows:
\begin{align}&\frac{a}{1-a}<\frac{b}{1-b}\\&\frac{a}{a-1}>\frac{b}{b-1}\\&\frac{a-1+1}{a-1}>\frac{b-1+1}{b-1}\\&1+\frac{1}{a-1}>1+\frac{1}{b-1}\\&\frac{1}{a-1}>\frac{1}{b-1}\end{align}
From $a<b$ it follows that $a-1<b-1<0$ and the inequality is thus satisfied.
A: If $a \lt b \lt 1$ then $(1 - a) \gt (1-b) \gt 0$. But then (inversion rule)
$\tag 1 0 \lt \frac{1}{1-a} \lt \frac{1}{1-b}$.
Multiplication of non-negative quantities preserves the inequality, so with (1) and
$\tag 2 0 \le a \lt b$
we get
$\tag 3 0 \le \frac{a}{1-a} \lt \frac{b}{1-b}$.
