Problem 6

For real numbers

x_1,x_2,y_1,y_2,z_1,z_2

satisfying

x_1>0

x_2>0

x_1y_1>z_1^2

, and

x_2y_2>z_2^2

, prove that

\frac{8}{(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le\frac1{x_1y_1-z_1^2}+\frac1{x_2y_2-z_2^2}

. Give necessary and sufficient conditions for equality.

Step 1 of 8: Name the positive determinants

a_i=x_iy_i-z_i^2>0\quad(i=1,2)

Detailed analysis

Set $a_1=x_1y_1-z_1^2$ and $a_2=x_2y_2-z_2^2$ . The hypotheses give $a_1,a_2>0$ . Also $y_1,y_2>0$ , because $x_i>0$ and $x_iy_i>z_i^2\ge0$ .