http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59017

《High-School Mathematics》(China Tianjin)No.3(1992) Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1x_2>0,x_1y_1>z_1^2, $ and $ x_2y_2>z_2^2,$ prove that \[(x_1+x_2)(y_1+y_2)\ge(z_1+z_2)^2+4\sqrt{(x_1y_1-z_1^2)(x_2y_2-z_2^2)}.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=500408&p=2811799#p2811799 Hungarian mathematics olympic Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1x_2>0,x_1y_1>z_1^2, $ and $ x_2y_2>z_2^2,$ prove that \[(x_1+x_2)(y_1+y_2)\ge(z_1+z_2)^2.\]

Rushil wrote: Given 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: \[ {8\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le{1\over x_1y_1-z_1^2}+{1\over x_2y_2-z_2^2}. \] Give necessary and sufficient conditions for equality. To strengthen it, I got Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1x_2>0,x_1y_1>z_1^2, $ and $ x_2y_2>z_2^2,$ prove that \[ {4\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le\frac{1}{\sqrt{(x_1y_1-z_1^2)(x_2y_2-z_2^2)}}\] $\Leftrightarrow (x_1+x_2)(y_1+y_2)\ge(z_1+z_2)^2+4\sqrt{(x_1y_1-z_1^2)(x_2y_2-z_2^2)}.$( you can see It's proof:here)

see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=498928

See also here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366742

Rushil wrote: Given 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: \[ {8\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le{1\over x_1y_1-z_1^2}+{1\over x_2y_2-z_2^2}. \]Give necessary and sufficient conditions for equality. Let $x_1, x_2,y _1,y_2$ be real numbers such that $x_1^2 + x_2^2 \le 1$. Prove the inequality$$(x_1y_1 + x_2y_2 - 1)^2 \ge (x_1^2 + x_2^2 - 1)(y_1^2 + y_2^2 -1)$$Austrian - Polish 1998