It is known that a, B, C and D satisfy a + B = 1, C + D = 1, AC + BD ＞ 1

It is known that a, B, C and D satisfy a + B = 1, C + D = 1, AC + BD ＞ 1

It is proved that: suppose that a, B, C and D are all nonnegative numbers, ∵ a + B = C + D = 1, ∵ (a + b) (c + D) = 1. ∵ AC + BD + BC + ad = 1 ≥ AC + BD

Given that a, B, C, D are real numbers, and a + B = C + D = 1, AC + BD > 1, it is proved that at least one of a, B, C, D is negative

Counter evidence
Suppose a, B, C and D are all positive numbers
Then 1 = (a + b) (c + D) = AC + AD + BC + BD
∴ac+bd=1-bc-ad＜1
It is contradictory to the condition AC + BD > 1
The hypothesis does not hold
At least one of a, B, C, D is negative

Given that a + B = C + D = 1, AC + BD > 1, we prove that at least one of a, B, C, D is negative

Counter evidence
Let a, B, C and d be all positive numbers
From a + B = C + D = 1, we can see that a, B, C and D are all numbers greater than 0 and less than 1
Let a = n, B = (1-N), 0

Given the real number zbcd, a + B = C + D = 1, AC + BD > 1, we prove that there is at least one negative number in ABCD

Suppose a, B, C and D are all nonnegative numbers
a+b=c+d=1
1=(a+b)(c+d)=ac+ad+bc+bd>1+ad+bc (ac+bd>1)
If a, B, C and D are all nonnegative numbers, then ad > = 0, BC > = 0
So 1 = (a + b) (c + D) > 1
contradiction
So at least one of a, B, C, D is negative