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 4 of 8: Cauchy–Schwarz gives a companion bound
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(c−a)2=[(b−a)+(c−b)]2≤2[(b−a)2+(c−b)2] ⟹ 3(c−a)2≤2[(b−a)2+(c−b)2+(c−a)2](c-a)^2=[(b-a)+(c-b)]^2\le 2\left[(b-a)^2+(c-b)^2\right]\ \Longrightarrow\ 3(c-a)^2\le 2\left[(b-a)^2+(c-b)^2+(c-a)^2\right]
Detailed analysis

By Cauchy–Schwarz (or (x+y)2≤2(x2+y2)(x+y)^2\le2(x^2+y^2)) with x=b−ax=b-a, y=c−by=c-b, we get (c−a)2=(x+y)2≤2(x2+y2)=2[(b−a)2+(c−b)2](c-a)^2=(x+y)^2\le2(x^2+y^2)=2\left[(b-a)^2+(c-b)^2\right]. Adding (c−a)2(c-a)^2 to both sides gives 3(c−a)2≤2[(b−a)2+(c−b)2+(c−a)2]3(c-a)^2\le2\left[(b-a)^2+(c-b)^2+(c-a)^2\right], again with equality iff b−a=c−bb-a=c-b.