IMO 2006 Shortlisted Problems

We first consider the cubic polynomial $$P(t)=tb\left(t^2-b^2\right)+bc\left(b^2-c^2\right)+ct\left(c^2-t^2\right)$$ It is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore $$P(t)=(b-c)(t-b)(t-c)(t+b+c)$$ since the cubic coefficient is $b-c$. The left-hand side of the proposed inequality can therefore be written in the form $$\left|ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)\right|=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)|$$ The problem comes down to finding the smallest number $M$ that satisfies the inequality $$\begin{array}{c}|(b-c)(a-b)(a-c)(a+b+c)|\le M\cdot {\left(a^2+b^2+c^2\right)}^2\end{array}\text{\hspace{1em}(1)}$$ Note that this expression is symmetric, and we can therefore assume $a\le b\le c$ without loss of generality. With this assumption, $$\begin{array}{c}|(a-b)(b-c)|=(b-a)(c-b)\le {\left(\frac{(b-a)+(c-b)}{2}\right)}^2=\frac{(c-a{)}^2}{4}\end{array}\text{\hspace{1em}(2)}$$ with equality if and only if $b-a=c-b$, i.e. $2b=a+c$. Also $${\left(\frac{(c-b)+(b-a)}{2}\right)}^2\le \frac{(c-b{)}^2+(b-a{)}^2}{2}$$ or equivalently, $$\begin{array}{c}3(c-a{)}^2\le 2\cdot \left[(b-a{)}^2+(c-b{)}^2+(c-a{)}^2\right]\end{array}\text{\hspace{1em}(3)}$$ again with equality only for $2b=a+c$. From (2) and (3) we get $$\begin{array}{rl}& |(b-c)(a-b)(a-c)(a+b+c)|\\ \le & \frac{1}{4}\cdot \left|(c-a{)}^3(a+b+c)\right|\\ =& \frac{1}{4}\cdot \sqrt{(c-a{)}^6(a+b+c{)}^2}\\ \le & \frac{1}{4}\cdot \sqrt{{\left(\frac{2\cdot \left[(b-a{)}^2+(c-b{)}^2+(c-a{)}^2\right]}{3}\right)}^3\cdot (a+b+c{)}^2}\\ =& \frac{\sqrt{2}}{2}\cdot {\left(\sqrt[4]{{\left(\frac{(b-a{)}^2+(c-b{)}^2+(c-a{)}^2}{3}\right)}^3\cdot (a+b+c{)}^2}\right)}^2\end{array}$$ By the weighted AM-GM inequality this estimate continues as follows: $$\begin{array}{rl}& |(b-c)(a-b)(a-c)(a+b+c)|\\ \le & \frac{\sqrt{2}}{2}\cdot {\left(\frac{(b-a{)}^2+(c-b{)}^2+(c-a{)}^2+(a+b+c{)}^2}{4}\right)}^2\\ =& \frac{9\sqrt{2}}{32}\cdot {\left(a^2+b^2+c^2\right)}^2.\end{array}$$ We see that the inequality (1) is satisfied for $M=\frac{9}{32}\sqrt{2}$, with equality if and only if $2b=a+c$ and $$\frac{(b-a{)}^2+(c-b{)}^2+(c-a{)}^2}{3}=(a+b+c{)}^2.$$ Plugging $b=(a+c)/2$ into the last equation, we bring it to the equivalent form $$2(c-a{)}^2=9(a+c{)}^2.$$ The conditions for equality can now be restated as $$2b=a+c\text{ and }(c-a{)}^2=18b^2.$$ Setting $b=1$ yields $a=1-\frac{3}{2}\sqrt{2}$ and $c=1+\frac{3}{2}\sqrt{2}$. We see that $M=\frac{9}{32}\sqrt{2}$ is indeed the smallest constant satisfying the inequality, with equality for any triple ( $a,b,c$ ) proportional to ( $1-\frac{3}{2}\sqrt{2},1,1+\frac{3}{2}\sqrt{2}$ ), up to permutation.