Given that ab ≠ 0, the sufficient and necessary condition for a + B = 1 is A3 + B3 + ab-a2-b2 = 0

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It is known that a and B belong to R, and the sufficient and necessary condition for "a > 1 and b > 1" is "a + b > 2 and ab - (a + b) + 1 > 0"

A > 1, b > 1, that is, A-1 > 0, B-1 > 0, then A-1 + B-1 > 0, that is, a + b > 2 (A-1) (B-1) > 0, that is, ab-a-b + 1 > 0, then "a + b > 2 and ab - (a + b) + 1 > 0" can be deduced from "a > 1 and b > 1", now there is a + b > 2, and ab-a-b + 1 > 0, then (A-1) (B-1) > 0, and (A-1) + (B-1) > 0, then it must be

If a is greater than B, then what is the necessary and sufficient condition for 1 / a less than 1 / b

a>1,b>1,

A necessary and sufficient condition for a times B to be greater than 0

A and B must be real numbers of the same sign, and neither of them is 0

Given that ab ≠ 0, the sufficient and necessary condition for a + B = 1 is A3 + B3 + ab-a2-b2 = 0