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 6 of 8: A hidden identity ties

u

and

s

a^2+b^2+c^2

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In plain words

Expanding $u+s^2$ directly, every cross term $ab$ , $bc$ , $ca$ cancels between the squared differences and the squared sum, leaving a pure multiple of $a^2+b^2+c^2$ — this is exactly what lets a fixed budget of $a^2+b^2+c^2$ be redistributed between $u$ and $s$ .

u+s^2=(b-a)^2+(c-b)^2+(c-a)^2+(a+b+c)^2=3\left(a^2+b^2+c^2\right)

Detailed analysis

Direct expansion gives $(b-a)^2+(c-b)^2+(c-a)^2+(a+b+c)^2=3(a^2+b^2+c^2)$ (the cross terms cancel), i.e. $u+s^2=3(a^2+b^2+c^2)$ where $R:=a^2+b^2+c^2$ is exactly the quantity on the right-hand side of the original inequality.