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 $x>-b$

As @trula suggested you have to pay attention for the sign to multiply an inequality. Since we don't know the sign of denominator in this case we can subtract instead. $$\begin{align}\frac{1}{x+a}&\lt\frac{1}{x+b} \\ \frac{1}{x+b}-\frac{1}{x+a}&\gt 0 \\ \frac{a-b}{(x+a)(x+b)}&\gt 0 \\ \frac{1}{(x+a)(x+b)}&\gt 0 \end{align}$$
Now considering $a\gt b\gt 0$ (given) we can write the below table for sign of L.H.S.,
$$\begin{array}{c|c|c}\hline x\lt -a &-a\lt x\lt -b &-b\lt x\\\hline (+)&(-)&(+)\\\end{array}$$ Thus, the condition is true for $x\lt-a$ or $x\gt-b$.
