MathLabs

Problem 3

Determine the least real number MM such that the inequality ∣ab(a2−b2)+bc(b2−c2)+ca(c2−a2)∣≤M(a2+b2+c2)2\left| ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})\right|\leq M(a^{2}+b^{2}+c^{2})^{2} holds for all real numbers aa, bb and cc.
Step 2 of 8: Order the variables (the bound is symmetric)
In plain words

Swapping any two of aa, bb, cc only flips the sign of (b−c)(a−b)(a−c)(a+b+c)(b-c)(a-b)(a-c)(a+b+c), and both the absolute value and (a2+b2+c2)2(a^2+b^2+c^2)^2 ignore that sign, so we may fix a convenient order.

a≤b≤ca\le b\le c
Detailed analysis

The whole inequality is unchanged by any permutation of aa, bb, cc: the right side (a2+b2+c2)2(a^2+b^2+c^2)^2 is manifestly symmetric, and the left side changes only up to sign under a permutation of the factored form (b−c)(a−b)(a−c)(a+b+c)(b-c)(a-b)(a-c)(a+b+c), which the absolute value erases. Hence we may assume a≤b≤ca\le b\le c.