IMO 1984 · Problem 1

Prove that \(0 \le yz + zx + xy - 2xyz \le 7/27\), where \(x\), \(y\) and \(z\) are non-negative real numbers for which \(x+y+z=1\).

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

Solution

Write \(F=yz+zx+xy-2xyz\). Since \(0\le x,y,z\le1\), we have \[ F=xy(1-z)+yz(1-x)+zx\ge0. \]

For the upper bound, expanding and using \(x+y+z=1\) gives \[ (1-2x)(1-2y)(1-2z) =1-2(x+y+z)+4(xy+yz+zx)-8xyz =4F-1. \]

If one of \(x,y,z\) is at least \(1/2\), say \(x\ge1/2\), then \(y,z\le1/2\). Consequently, \[ (1-2x)(1-2y)(1-2z)\le0, \] so \[ F\le\frac14<\frac7{27}. \]

Otherwise, \(1-2x\), \(1-2y\) and \(1-2z\) are all positive, and their sum is \(1\). By the arithmetic–geometric mean inequality, \[ (1-2x)(1-2y)(1-2z) \le\left(\frac{(1-2x)+(1-2y)+(1-2z)}3\right)^3 =\frac1{27}. \] Thus \[ F\le\frac{1+1/27}{4}=\frac7{27}, \] as required.

The upper bound is attained exactly when \(x=y=z=1/3\). For the lower bound, the initial decomposition shows that \(F=0\) forces \(zx=0\); substituting \(x=0\) or \(z=0\) then shows that equality occurs exactly at the permutations of \((1,0,0)\).

The solution was checked against John Scholes' IMO solution archive (Kalva).