How to prove: A ^ 3 + B ^ 3 + C ^ 3 > = 3ABC a. B and C are positive numbers

How to prove: A ^ 3 + B ^ 3 + C ^ 3 > = 3ABC a. B and C are positive numbers

Given a ^ 3 + B ^ 3 + C ^ 3 = 3ABC, it is proved that a = b = C
A, B and C are not less than zero

A ^ 3 + B ^ 3 + C ^ 3-3abc = 0 (a + b) (a ^ 2-AB + B ^ 2) + C (C ^ 2-3ab) = 0 (a + b) (a ^ 2-AB + B ^ 2) + C (C ^ 2-AB + B ^ 2) + C (C ^ 2-3ab + A ^ 2-AB + B ^ 2-A ^ 2 + ab-b ^ 2) = 0 (a + b) (a ^ 2-AB + B ^ 2) + C [(C ^ 2-A ^ 2-2ab-b ^ 2) + (a ^ 2-AB + B ^ 2 = 0

How to prove (a + B + C) ^ 3 > = 3ABC
Such as the title

A ^ 3 + B ^ 3 + C ^ 3-3abc = (a + B + C) (a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac)
Because a B C is nonnegative, a + B + C > = 0 a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac = 1 / 2 [(a-b) ^ 2]+
(B-C) ^ 2 + (C-A) ^ 2] > = 0, so left > = 0