Given 1 > a > b > C > 0, we prove that (1-A) · (1-B) · (1-C) is greater than or equal to 8abc It is known that the three sides of a triangle with perimeter 1 are A.B., C | Math enthusiast | Mathematical art

Given 1 > a > b > C > 0, we prove that (1-A) · (1-B) · (1-C) is greater than or equal to 8abc It is known that the three sides of a triangle with perimeter 1 are A.B., C

（1-a）·（1-b）·（1-c）=（a+b)*(b+c)*(a+c)=a^2b+a^bc+ab^2+abc+abc+ac^2+b^2c+bc^2
a(b*2+c*2)+b(a*2+c*2)+c(a*2+b*2)+2abc
Because (B-C) * 2 > = 0, so b * 2 + C * 2 > = 2BC, because 0 = 2abc, C (b * 2 + A * 2) > = 2abc, so (1-A) · (1-B) · (1-C) > = 8abc

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