Range of values of x for which $a\gt b \rightarrow \frac{1}{x+a} \lt \frac{1}{x+b}$ I am working through a pure maths book as a hobby. I have come to inequalities and am struggling with this question.
Given that $a$ and $b$ are positive, state the range of values of x for which the following statement is true:$a\gt b \rightarrow \frac{1}{x+a} \lt \frac{1}{x+b}$
I have said, $\frac{1}{x+a} \lt \frac{1}{x+b} \rightarrow 1<\frac{x+a}{x+b} \rightarrow \frac{x+a}{x+b} \gt 1 \rightarrow x+a\gt x+b$
So the original statement can be expressed as
$a\gt b \rightarrow  x+a\gt x+b$, which is true for all x.
But the book says the answer is: $x\lt -a$ or $x>-b$

Multiplying  the inequality by $x+a$ and conserving the < sign is only right if $x+a>0$ if $x+a<0$ you get $1>\frac{x+a}{x+b}$
multiplying inequality you always have to take in account the sign of the factor.
