How does one manipulate rational absolute inequalities?
For example, I want to transform the absolute value inequality $$|x-3|<1,$$ to $$\frac{|x+3|}{5x^2}<A,$$ for some number A, and using $$|x-3|<1$$ as a constraint or starting point, and to find an upper and lower bound on the latter term and not sure what to do with the denominator and changing inequality direction.
I can expand the absolute value inequality as follows: $$|x-3|<1 \implies -1<x-3<1 \implies 2<x<4 \implies 5<x+3<7. $$ How do I introduce the $5x^2$ term in the denominator? Dividing all three sides by $5x^2$ would add a variable to the numbers and possibly change signs. Not sure where to go from here.
For context, this question is not for homework but in reference to bounding terms for epsilon-delta limit proofs. I know you can bound the numerator and denominator separately but I really would like to know how manipulate the quotient together as I have forgotten some elementary algebra.
Thanks!
Edit: More info: If $$|x-3|<1, $$ then $$|x-3|<1 \implies -1<x-3<1 \implies 2<x<4 \implies 5<x+3<7$$ for the numerator and $$\frac{1}{4}<\frac{1}{x}<\frac{1}{2} \implies \frac{1}{16}<\frac{1}{x^2}<\frac{1}{4} \implies \frac{1}{80}<\frac{1}{5x^2}<\frac{1}{20}$$ for the denominator.
Dividing produces $$\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies -\frac{7}{20}<\frac{5}{80}<\frac{x+3}{5x^2}<\frac{7}{20} \implies \frac{|x+3|}{5x^2}<\frac{7}{20}.$$
The questions I have are (a) can the numerator and denominator be separated and recombined as above? If not, how to I transform $$|x-3|<1$$ to $$\frac{|x+3|}{5x^2}<\frac{7}{20}$$ without splitting the fraction?
Does that help?
The best way to résove this inequality is to to study the function f(x)=|x+3|/(5x²) for x in [2,4] you find that 0<x<1/20enter image description here