Find all positive real solutions to:
(x12 - x3x5)(x22 - x3x5) ≤ 0
(x22 - x4x1)(x32 - x4x1) ≤ 0
(x32 - x5x2)(x42 - x5x2) ≤ 0
(x42 - x1x3)(x52 - x1x3) ≤ 0
(x52 - x2x4)(x12 - x2x4) ≤ 0
Solution
Answer: x1 = x2 = x3 = x4 = x5.
The difficulty with this problem is that it has more information than we need. There is a neat solution in Greitzer which shows that all we need is the sum of the 5 inequalities, because one can rewrite that as (x1x2 - x1x4)2 + (x2x3 - x2x5)2 + ... + (x5x1 - x5x3)2 + (x1x3 - x1x5)2 + ... + (x5x2 - x5x4)2 ≤ 0. The difficulty is how one ever dreams up such an idea!
The more plodding solution is to break the symmetry by taking x1 as the largest. If the second largest is x2, then the first inequality tells us that x12 or x22 = x3x5. But if x3 and x5 are unequal, then the larger would exceed x1 or x2. Contradiction. Hence x3 = x5 and also equals x2 or x1. If they equal x1, then they would also equal x2 (by definition of x2), so in any case they must equal x2. Now the second inequality gives x2 = x1x4. So either all the numbers are equal, or x1 > x2 = x3 = x5 > x4. But in the second case the last inequality is violated. So the only solution is all numbers equal.
If the second largest is x5, then we can use the last inequality to deduce that x2 = x4 = x5 and proceed as before.
If the second largest is x3, then the fourth inequality gives that x1 = x3 = x5 or x1 = x3 = x4. In the first case, x5 is the second largest and we are home already. In the second case, the third inequality gives x32 = x2x5 and hence x3 = x2 = x5 (or one of x2, x5 would be larger than the second largest). So x5 is the second largest and we are home.
Finally, if the second largest is x4, then the second inequality gives x1 = x2 = x4 or x1 = x3 = x4. Either way, we have a case already covered and so the numbers are all equal.
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.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 1998
Last updated/corrected 14 Mar 03