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 8 of 8: Conclude the sharp constant and exhibit equality

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|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|\le\frac{9\sqrt2}{32}(a^2+b^2+c^2)^2,\qquad M=\frac{9\sqrt2}{32}

Detailed analysis

Combining Steps 5 and 7, $|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|=|(b-c)(a-b)(a-c)(a+b+c)|\le\frac14\left(\frac{2u}{3}\right)^{3/2}s\le\frac{9\sqrt2}{32}(a^2+b^2+c^2)^2$ , so $M=\frac{9\sqrt2}{32}$ works. Equality needs $2b=a+c$ (Steps 3–4) and $\frac u3=s^2$ (Step 7); one checks that $(a,b,c)=(2-3\sqrt2,\,2,\,2+3\sqrt2)$ satisfies both and makes the two sides equal, so $\frac{9\sqrt2}{32}$ is optimal: $M=\frac{9\sqrt2}{32}$ .