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 3 of 8: Apply weighted AM-GM
Previous stepNext step
2a1a2≤a1y2y1+a2y1y22\sqrt{a_1a_2}\le a_1\frac{y_2}{y_1}+a_2\frac{y_1}{y_2}
Detailed analysis

Apply AM-GM to a1y2/y1a_1y_2/y_1 and a2y1/y2a_2y_1/y_2. Their product is a1a2a_1a_2, so the displayed inequality follows.