Question

If `a` and `b` are integers with `a > b`, what is the smallest possible positive value of `\frac{a+b}{a-b} + \frac{a-b}{a+b}`?

Question from AoPS.
Subject: Algebra
Focus: Quadratic Inequalities

Answer 1

Smallest possible value of `(a+b)/(a-b)+(a-b)/(a+b)` is `2`

Explanation:

`(a+b)/(a-b)+(a-b)/(a+b)`

= `((a+b)^2+(a-b)^2)/(a^2-b^2)`

= `(2a^2+2b^2)/(a^2-b^2)`

= `2+(4b^2)/(a^2-b^2)`

As in `(4b^2)/(a^2-b^2)`, both numerator and denominator are always positive, its lease value will be `0`, when `b=0`

and hence smallest possible value of `(a+b)/(a-b)+(a-b)/(a+b)` is `2`