a, b, c and d are four positive real numbers such that abcd=1, what is the minimum value of (1+a)(1+b)(1+c)(1+d)?
As Bunuel said, abcd = 1 implies that either the numbers are equal to 1 or there are pairs of reciprocals e.g. (1, 1, 1, 1) or (1, 1, 2, 1/2) or (3, 1/3, 4, 1/4) etc.
If a and b are 1 and 1,
(1+a)(1+b) = (1+1)(1+1) = 4
If a and b are 2 and 1/2,
(1+a)(1+b) = (1+2)(1+1/2) = 9/2 = 4.5
If a and b are 3 and 1/3,
(1+a)(1+b) = (1+3)(1+1/3) = 16/3 = 5.3
As you keep taking higher reciprocals, the value of (1+a)(1+b) keeps increasing.
So taking reciprocals is a bad idea and all numbers must be 1 giving us the minimum value of 16.
Anyway, in any minimum-maximum question, it is a good idea to check on equality. Often, the point of equality is a transition point.
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