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 4 of 8: Use a nonnegative square
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2z1z2≤z12y2y1+z22y1y22z_1z_2\le z_1^2\frac{y_2}{y_1}+z_2^2\frac{y_1}{y_2}
Detailed analysis

Since y1,y2>0y_1,y_2>0, the square y1y2(z1/y1−z2/y2)2y_1y_2(z_1/y_1-z_2/y_2)^2 is nonnegative. Expanding it gives the displayed inequality.