Let the positive number be x. So we have x > 0

we want x + 1/x to be a minimum.

let y = x + 1/x

=> y' = 1 - 1/x^2

for min, set y' = 0

=> 1 - 1/x^2 = 0

=> 1/x^2 = 1

=> x^2 = 1

=> x = +/- 1

=> x = 1, since x > 0

you can check +1 and -1 using the second derivative test if you want to be sure of this, or graph x + 1/x