1999 IMO Problems/Problem 2
Problem
(Marcin Kuczma, Poland) Let be a fixed integer.
- (a) Find the least constant such that for all nonnegative real numbers ,
- (b) Determine when equality occurs for this value of .
Solution
The answer is , and equality holds exactly when two of the are equal to each other and all the other are zero. We prove this by induction on the number of nonzero .
First, suppose that at most two of the , say and , are nonzero. Then the left-hand side of the desired inequality becomes and the right-hand side becomes . By AM-GM, with equality exactly when , as desired.
Now, suppose that our statement holds when at most of the are equal to zero. Suppose now that of the are equal to zero, for . Without loss of generality, let these be . We define and for convenience, we will denote . We wish to show that by replacing the with the , we increase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.
We note that If we replace with , then becomes but none of the other terms change. Since , it follows that we have strictly increased the right-hand side of the equation, i.e., By inductive hypothesis, and by our choice of , Hence the problem's inequality holds by induction, and is strict when there are more than two nonzero , as desired.
Solution 2 (elegant)
I claim that . Let and . Note that Equality holds in the first ineq when all but two of the are zero, and this reduces to the case which we can easily show to be equivalent to , so . That is, equality holds when two are the same and the rest are zero.
Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
1999 IMO (Problems) • Resources | ||
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