If $a$ and $b$ are integers with $a > b$, what is the smallest possible positive value of \(\frac{a+b}{a-b} + \frac{a-b}{a+b}\)?
+128408
(a + b) / (a -b) + (a - b) / (a + b) → [ (a + b)^2 + ( a - b)^2 ] / [ a^2 - b^2] →
[ a ^ 2 + 2ab + b^2 + a^2 - 2ab + b^2) / [ a^2 - b^2] →
[ 2a ^2 + 2 b^2] / [ a^2 - b^2] →
2 [ a^2 + b^2] / [a^2 - b^2 ] →
This expression will have the smallest possible value when the numerator of a^2 + b^2 = the denominator of a^2 - b^2
Note that when a = any integer > 0 and b = 0 the numerator of a^2 + b^2 = the denominator of a^2 - b^2 = 1
So....the smallest positive value is 2
