Problem 3

Determine the least real number

M

such that the inequality

\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

a

b

and

c

Step 3 of 8: AM–GM on the two short gaps

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In plain words

Since $a\le b\le c$ , both $b-a$ and $c-b$ are nonnegative, so their product is at most the square of their average — a textbook AM–GM.

|(a-b)(b-c)|\le\left(\frac{(b-a)+(c-b)}{2}\right)^2=\frac{(c-a)^2}{4}

Detailed analysis

By AM–GM for the nonnegative numbers $b-a$ and $c-b$ , $(b-a)(c-b)\le\left(\frac{(b-a)+(c-b)}{2}\right)^2=\frac{(c-a)^2}{4}$ , and since $(a-b)(b-c)=(b-a)(c-b)\ge0$ this equals $|(a-b)(b-c)|$ . Equality holds iff $b-a=c-b$ , i.e. $2b=a+c$ .