MathLabs

Problem 6

For real numbers x1,x2,y1,y2,z1,z2x_1,x_2,y_1,y_2,z_1,z_2 satisfying x1>0x_1>0, x2>0x_2>0, x1y1>z12x_1y_1>z_1^2, and x2y2>z22x_2y_2>z_2^2, prove that 8(x1+x2)(y1+y2)−(z1+z2)2≤1x1y1−z12+1x2y2−z22\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 6 of 8: Multiply the estimates
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4a1a2≤(a1+a2)(x1y2+x2y1−2z1z2)4a_1a_2\le(a_1+a_2)(x_1y_2+x_2y_1-2z_1z_2)
Detailed analysis

Multiply the cross-term inequality by 2a1a22\sqrt{a_1a_2} and use 2a1a2≤a1+a22\sqrt{a_1a_2}\le a_1+a_2. This gives the displayed inequality.