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 7 of 8: A weighted AM–GM turns the fixed budget into the sharp bound

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In plain words

Since $u/3$ appears three times and $s^2$ once in the fixed sum $u+s^2=3R$ , weighting the AM–GM $3$ -to- $1$ matches exactly the powers $\left(\frac u3\right)^{3}s^2$ that show up in $\left(\frac{2u}{3}\right)^{3/2}s$ .

\frac{\frac u3+\frac u3+\frac u3+s^2}{4}\ge\sqrt[4]{\left(\frac u3\right)^3s^2}\ \Longrightarrow\ \left(\frac{2u}{3}\right)^{3/2}s\le\frac{9\sqrt2}{8}\left(a^2+b^2+c^2\right)^2

Detailed analysis

Apply AM–GM to the four nonnegative numbers $\frac u3$ , $\frac u3$ , $\frac u3$ , $s^2$ , whose sum is $u+s^2=3R$ by Step 6: $\frac{3\cdot\frac u3+s^2}{4}\ge\sqrt[4]{\left(\frac u3\right)^3s^2}$ gives $\left(\frac u3\right)^3s^2\le\left(\frac{3R}{4}\right)^4$ . Multiplying by $8$ and taking square roots turns $\left(\frac{2u}{3}\right)^3s^2=8\left(\frac u3\right)^3s^2$ into $\left(\frac{2u}{3}\right)^{3/2}s\le\sqrt{8}\left(\frac{3R}{4}\right)^2=\frac{9\sqrt2}{8}R^2$ .