Given real numbers x1, x2, y1, y2, z1, z2, satisfying x1 > 0, x2 > 0, x1y1 > z12, and x2y2 > z22, prove that:
8/((x1 + x2)(y1 + y2) - (z1 + z2)2) <= 1/(x1y1 - z12) + 1/(x2y2 - z22).
Give necessary and sufficient conditions for equality.
Solution
Let a1 = x1y1 - z12 and a2 = x2y2 - z22. We apply the arithmetic/geometric mean result 3 times:
(1) to a12, a22, giving 2a1a2 <= a12 + a22;
(2) to a1, a2, giving Ö(a1a2) <= (a1 + a2)/2;
(3) to a1y2/y1, a2y1/y2, giving Ö(a1a2) <= (a1y2/y1 + a2y1/y2)/2;
We also use (z1/y1 - z2/y2)2 >= 0. Now x1y1 > z12 >= 0, and x1 > 0, so y1 > 0. Similarly, y2 > 0. So:
(4) y1y2(z1/y1 - z2/y2)2 >= 0, and hence z12y2/y1 + z22y1/y2 >= 2z1z2.
Using (3) and (4) gives 2Ö(a1a2) <= (x1y2 + x2y1) - (z12y2/y1 + z22y1/y2) <= (x1y2 + x2y1 - 2z1z2).
Multiplying by (2) gives: 4a1a2 <= (a1 + a2)(x1y2 + x2y1 - 2z1z2).
Adding (1) and 2a1a2 gives: 8a1a2 <= (a1 + a2)2 + (a1 + a2)(x1y2 + x2y1 - 2z1z2) = a(a1 + a2), where a = (x1 + x2)(y1 + y2) - (z1 + z2)2. Dividing by a1a2a gives the required inequality.
Equality requires a1 = a2 from (1), y1 = y2 from (2), z1 = z2 from (3), and hence x1 = x2. Conversely, it is easy to see that these conditions are sufficient for equality.