1999 IMO Problems/Problem 2
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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,
![\begin{align*}x_a x_b(x_a^2 + x_b^2) = 1/2 \left[ \sqrt{(2x_ax_b)(x_a^2 + x_b^2)} \right]^2&\le \frac{1}{2} \left( \frac{...](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/c/6/bc62d256d545ec1cbe74dbd6c0ea2e9ff450ba24.gif)
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. 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
