Difference between revisions of "2000 USAMO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | Call a real-valued function <math>f</math> ''very convex'' if | + | Call a real-valued [[function]] <math>f</math> ''very convex'' if |
<cmath>\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|</cmath> | <cmath>\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|</cmath> | ||
− | holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very | + | holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very convex function exists. |
== Solution == | == Solution == |
Latest revision as of 00:45, 9 July 2017
Problem
Call a real-valued function very convex if
holds for all real numbers and
. Prove that no very convex function exists.
Solution
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,
A straightforward induction shows that:
for all nonnegative integers
Using a similar line of reasoning as above,
Therefore, for every nonnegative integer
Now, we choose
large enough such that
This is always possible because
and
are fixed for any particular function
It follows that:
However, by the very convex condition,
This is a contradiction. It follows that no very convex function exists.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.