IMO 1969 · Problem 6

Prove that for all real numbers \(x_1,x_2,y_1,y_2,z_1,z_2\), with \(x_1>0,\ x_2>0,\ x_1y_1-z_1^2>0,\ x_2y_2-z_2^2>0\), the inequality \[ \frac{8}{(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2} \leq \frac{1}{x_1y_1-z_1^2}+\frac{1}{x_2y_2-z_2^2} \] is satisfied. Give necessary and sufficient conditions for equality.

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

Answer

Equality holds if and only if \[ x_1=x_2,\qquad y_1=y_2,\qquad z_1=z_2. \]

Solution

For \(i=1,2\), put \[ u_i=y_i-\frac{z_i^2}{x_i}>0, \qquad d_i=x_iy_i-z_i^2=x_i u_i>0. \] Write \[ D=(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2. \] Substitution and expansion give \[ D=(x_1+x_2)(u_1+u_2) +\frac{(x_2z_1-x_1z_2)^2}{x_1x_2}. \] Consequently, by the arithmetic–geometric mean inequality, \[ D\geq (x_1+x_2)(u_1+u_2) \geq 4\sqrt{x_1x_2u_1u_2} =4\sqrt{d_1d_2}>0. \] Applying the same inequality to \(1/d_1\) and \(1/d_2\), we obtain \[ \frac{8}{D} \leq \frac{2}{\sqrt{d_1d_2}} \leq \frac1{d_1}+\frac1{d_2}, \] which proves the required inequality.

Suppose equality holds. Then equality must hold throughout this chain, and hence throughout the estimate for \(D\). Equality in the two arithmetic–geometric mean estimates \[ x_1+x_2\geq2\sqrt{x_1x_2}, \qquad u_1+u_2\geq2\sqrt{u_1u_2} \] forces \(x_1=x_2\) and \(u_1=u_2\). The squared term must also vanish, so \(x_2z_1=x_1z_2\). Since \(x_1=x_2>0\), this gives \(z_1=z_2\). Finally, \[ y_i=u_i+\frac{z_i^2}{x_i} \] implies \(y_1=y_2\).

Conversely, if \(x_1=x_2=x\), \(y_1=y_2=y\), and \(z_1=z_2=z\), then \(d=xy-z^2>0\), and both sides of the proposed inequality equal \(2/d\). Thus the stated conditions are both necessary and sufficient.

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