Question

Solve the following inequality:
$\frac{1}{\sqrt{1+x}}>\frac{1}{2-x}.$

Answer 1

$\frac{1}{\sqrt{1+x}}>\frac{1}{2-x}$

Squaring both sides

$\frac{1}{x+1}>\frac{1}{(x-2{)}^2}$

$\frac{1}{x+1}-\frac{1}{(x-2{)}^2}>0$

$\frac{(x-2{)}^2-(x+1)}{(x+1)(x-2{)}^2}>0$

$\frac{x^2-3x+3}{(x+1)(x-2{)}^2}>0$

$\therefore \frac{1}{(x+1)(x-2{)}^2}>0$

$\therefore x\in (-1,\infty )-\left\{2\right\}$