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

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Subject: Algebra
Focus: Quadratic Inequalities

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Answer 1 · 2018-07-31T05:53:47

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

$(a+b)/(a-b)+(a-b)/(a+b)$

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

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

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

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 $2$