IMO 1969

Problem B3

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 √(a1a2) ≤ (a1 + a2)/2;

(3) to a1y2/y1, a2y1/y2, giving √(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 2√(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.

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.