Using (c-a) = -(a-b) - (b-c) and wlog, assuming a>=b>=c, one rewrites the original expression S as: S = [a²b-c²a](a-b) + [b²c-c²a](b-c). Now as a²b>=abc & -ac²>=-abc, one substitutes and factors obtaining: S >= ac(b-c)(a-b) + bc(b-a)(b-c) or S >= (ac - bc)(b - c)(a - b) when S >=0 since each term in parentheses is non-negative.
Thank you for posting your solution. You did not use the triangular condition. But that condition is indeed necessary. Here is a counterexample: a=1, b=2 and c=10. The inequality does not hold. When a=10, b=2 and c=1 (the assumptions in your proof), it holds! This is somewhat subtle: The problem is ""cyclic" but not completely symmetric, i.e., it contains the term a^2 b (a-b) but not b^2 a (b-a). Regarding the orders of a, b, and c, there are total 6 permutations, in two cyclic groups: group 1: a >=b >=c, b>=c>=a, or c>=a>=b. Your proof covers these three cases. You do not need the triangular condition. group 2: b>=a>=c, a>=c>=b, or c>=b>=a. These three cases are the tricky ones. I took a quick look but could not figure it out. Let me know if you solve the cases! Again, appreciate your comment!
Using (c-a) = -(a-b) - (b-c) and wlog, assuming a>=b>=c, one rewrites the original expression S as:
S = [a²b-c²a](a-b) + [b²c-c²a](b-c). Now as a²b>=abc & -ac²>=-abc, one substitutes and factors obtaining:
S >= ac(b-c)(a-b) + bc(b-a)(b-c) or
S >= (ac - bc)(b - c)(a - b) when S >=0 since each term in parentheses is non-negative.
Thank you for posting your solution. You did not use the triangular condition. But that condition is indeed necessary. Here is a counterexample: a=1, b=2 and c=10. The inequality does not hold. When a=10, b=2 and c=1 (the assumptions in your proof), it holds! This is somewhat subtle: The problem is ""cyclic" but not completely symmetric, i.e., it contains the term a^2 b (a-b) but not b^2 a (b-a).
Regarding the orders of a, b, and c, there are total 6 permutations, in two cyclic groups:
group 1: a >=b >=c, b>=c>=a, or c>=a>=b. Your proof covers these three cases. You do not need the triangular condition.
group 2: b>=a>=c, a>=c>=b, or c>=b>=a. These three cases are the tricky ones. I took a quick look but could not figure it out. Let me know if you solve the cases!
Again, appreciate your comment!