How to prove that ab + BC + AC is greater than or equal to 8abc The original problem is that a, B and C are all positive numbers, and a + B + C = 1. Prove that: (1-A) (1-B) (1-C) is greater than or equal to 8abc

How to prove that ab + BC + AC is greater than or equal to 8abc The original problem is that a, B and C are all positive numbers, and a + B + C = 1. Prove that: (1-A) (1-B) (1-C) is greater than or equal to 8abc

If a, B and C are any positive numbers, this inequality does not hold
For example, if a = b = C = 1, then AB + BC + AC = 3

a. B, C are positive integers, a > b, the square of a minus AB minus AC plus BC equals 7, then a minus C equals what
A. - 2 b. - 1 c.0 D.2 is available

A^2-AB-AC+BC=7
（a^2-ac)-(ab-bc)=7
(a-b)(a-c)=7=1*7=(-1)(-7)
a-c=7/(a-b)
a> B, and a, B, C are all positive integers, so A-B > = 1
So A-B = 1, or A-B = 7, so
A-c = 7, or 1

If a minus B equals 2 and a minus C equals half, find the value of a plus B plus C minus AB minus AC minus BC

a-b=2, a-c=½∴b-c=(a-c)-(a-b)=-1.5a²+b²+c²-ab-ac-bc=½﹙2a²+2b²+2c²-2ab-2ac-2bc)=½×[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]=...