2000 USAMO Problems/Problem 1
Contents
Problem
Call a real-valued function very convex if
holds for all real numbers and . Prove that no very convex function exists.
Solution 1
Let , and substitute . Then a function is very convex if , or rearranging,
Let , which is the slope of the secant between . Let be arbitrarily small; then it follows that , . Summing these inequalities yields . As (but , so is still arbitrarily small), we have . This implies that in the vicinity of any , the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.
Solution 2
Suppose, for the sake of contradiction, that there exists a very convex function Notice that is convex if and only if is convex, where is a constant. Thus, we may set for convenience.
Suppose that and By the very convex condition, n.n,nn > \frac{A+B}{4} - 1.ABf.$ It follows that: However, by the very convex condition, This is a contradiction. It follows that there exists no very convex function.
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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