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#