1977 USAMO Problems/Problem 5
If are positive numbers bounded by and , i.e, if they lie in , prove that and determine when there is equality.
Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but . Then the expression on the LHS has the form , where and are fixed. But this is convex. That is to say, as increases if first decreases, then increases. So its maximum must occur at or . This is true for each variable.
Suppose all five are or all five are , then the LHS is 25, so the inequality is true and strict unless . If four are and one is , then the LHS is . Similarly if four are and one is . If three are and two are , then the LHS is . Similarly if three are and two are .
with equality iff , so if , then three of one and two of the other gives a larger LHS than four of one and one of the other. Finally, we note that the RHS is in fact , so the inequality is true with equality iff either (1) or (2) three of are and two are or vice versa.
|1977 USAMO (Problems • Resources)|
|1 • 2 • 3 • 4 • 5|
|All USAMO Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.