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 8 of 8: Divide and characterize equality

Previous step Next step

\frac8A\le\frac1{a_1}+\frac1{a_2},\quad x_1=x_2,\ y_1=y_2,\ z_1=z_2

Detailed analysis

Step 5 gives $C=x_1y_2+x_2y_1-2z_1z_2\ge2\sqrt{a_1a_2}$ , so $A=C+a_1+a_2\ge a_1+a_2+2\sqrt{a_1a_2}>0$ . Thus $Aa_1a_2>0$ , and division gives the required inequality. Equality in all steps requires $a_1=a_2$ , $y_1=y_2$ , and $z_1/y_1=z_2/y_2$ , hence $z_1=z_2$ and then $x_1=x_2$ . Conversely these equalities make every step an equality.