1994 USAMO Problems/Problem 4
Problem 4
Let be a sequence of positive real numbers satisfying
for all
. Prove that, for all
Solution
Since each is positive, by Muirhead's inequality,
. Now we claim that
, giving
works, but we set the base case
, which gives
. Now assume that it works for
.
By our assumption, now we must prove that, for
case,
, which is clearly true for
. So we are done.
See Also
1994 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |