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 7 of 8: Build the left denominator

Previous step Next step

4a_1a_2\le(a_1+a_2)C,\quad 4a_1a_2\le(a_1+a_2)^2\quad\Longrightarrow\quad 8a_1a_2\le(a_1+a_2)(C+a_1+a_2)=A(a_1+a_2),\quad C=x_1y_2+x_2y_1-2z_1z_2

Detailed analysis

Combine Step 6 with $4a_1a_2\le(a_1+a_2)^2$ , which is equivalent to $(a_1-a_2)^2\ge0$ . With $C=x_1y_2+x_2y_1-2z_1z_2$ and $A=C+a_1+a_2$ , Step 6 gives $4a_1a_2\le(a_1+a_2)C$ ; adding the two inequalities yields $8a_1a_2\le A(a_1+a_2)$ .