Question 6
Under the conditions $x_1,x_2>0$, $x_1y_1>z_1^2$, and $x_2y_2>z_2^2$, prove the inequality
$$\frac8{(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le\frac1{x_1y_1-z_1^2}+\frac1{x_2y_2-z_2^2}.$$
$$\frac8{(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le\frac1{x_1y_1-z_1^2}+\frac1{x_2y_2-z_2^2}.$$