bc(b^3-c^3) +ca(c^3+a^3) + ab(a^3-b^3)
Answers
Answer:
Note: I got cross-eyed after awhile, so there may be arithmetic errors! There may well be easier ways to do this… LOL
The desired statement would be true if
2(a^3 + b^3 + c^3) - ab(a+b) - bc(b+c) - ca(c+a) >= 0
This can be re-expressed as
2a^3 - a^2(b+c) + 2b^3 - b^2(a+c) + 2c^3 - c^2(a+b) >= 0
a^2(2a - (b+c)) + b^2(2b - (a+c)) + c^2(2c - (a+b)) >= 0
Consider the case where a = b = c. In each term, within the outermost parentheses, the expression equals zero.
Consider the case where two of the variables are equal, say a = b. The inequality would then reduce to
2a^2(a-c) + c^2(2c - 2a) >= 0
2a^2(a-c) + 2c^2(c-a) >= 0
2(a-c)(a^2 - c^2) >= 0
Regardless of whether a > c, a < c, (or a = c, although this case was already covered in considering a = b = c), the above inequality is true, as the factors in parentheses are either both zero, both positive, or both negative.
A similar argument applies to considering the cases were a = c or b = c, because of the symmetry of the expression.
The remaining case is that a, b, and c are distinct. There are six possible relationships between them:
a > b > c
a > c > b
b > a > c
b > c > a
c > a > b
c > b > a
recall:
a^2(2a - (b+c)) + b^2(2b - (a+c)) + c^2(2c - (a+b)) >= 0
Consider the case a > b > c.
Examing the first term, outermost parentheses factor:
2a - (b+c) = (a-b) + (a-c) > 2(a-b), since b > c,
2(a-b) < 2a - (b+c) < 2a
Examining the second term, outermost parentheses factor:
2b - (a+c) = (b-a) + (b-c) < 2(b-c), since a > c
2b - (a+c) < -2(c-b)
Examining the the third term, outermost parentheses factor:
2c - (a+b) = (c-a) + (c-b) < 2(c-b), since a > b
2c - (a+b) < -2(b-c)
Put it together, examining “worst case scenario”
a^2(factor greater than 2(a-b)) - b^2(factor less than 2(c-b)) - c^2 (factor less than 2(b-c))
Since a > b > c, we need only confirm that
(a-b) - (c-b) - (b-c) >= 0
a - b >= 0, which was given (a > b).
By symmetry, this will work for the other five cases.
Hope this helps and doesn’t mislead or confuse you.