IMO 2006 · Problem 3

Determine the least real number \(M\) such that the inequality \[ \left|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)\right| \le M(a^2+b^2+c^2)^2 \] holds for all real numbers \(a\), \(b\) and \(c\).

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{M=\frac{9\sqrt2}{32}}. \]

Solution

Write \[ F=ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2),\qquad R=a^2+b^2+c^2. \] If \(R=0\), then \(a=b=c=0\), and the inequality holds. Henceforth assume \(R>0\).

Set \[ s=a+b+c,\qquad x=a-\frac{s}{3},\quad y=b-\frac{s}{3},\quad z=c-\frac{s}{3}, \qquad q=x^2+y^2+z^2. \] Then \[ x+y+z=0,\qquad R=\frac{s^2}{3}+q. \] Factoring the expression gives \[ F=(a+b+c)(a-b)(a-c)(b-c) =s(x-y)(x-z)(y-z). \]

For \(x+y+z=0\), we have the algebraic identity \[ (x-y)^2(x-z)^2(y-z)^2+27x^2y^2z^2 =\frac12(x^2+y^2+z^2)^3. \] Indeed, substituting \(z=-x-y\), the left-hand side becomes \[ (x-y)^2(2x+y)^2(x+2y)^2+27x^2y^2(x+y)^2 =4(x^2+xy+y^2)^3, \] which is also the right-hand side. Consequently, \[ F^2\le \frac12s^2q^3. \]

Applying AM–GM to the four nonnegative numbers \(s^2,q,q,q\), we obtain \[ s^2q^3 \le \left(\frac{s^2+3q}{4}\right)^4 =\left(\frac{3R}{4}\right)^4. \] Therefore \[ F^2\le \frac{81}{512}R^4, \] and taking square roots yields \[ |F|\le \frac{9\sqrt2}{32}(a^2+b^2+c^2)^2. \]

To prove that the constant is least possible, take \[ (a,b,c)=(\sqrt2+3,\sqrt2,\sqrt2-3). \] Here \(s=3\sqrt2\), \((x,y,z)=(3,0,-3)\), and \(q=18=s^2\). Thus equality holds in both inequalities above. Since \(R=24>0\), no smaller constant can satisfy the required inequality.

Hence the least value is \(\boxed{\frac{9\sqrt2}{32}}\).

The solution was checked against the official IMO shortlist solutions.