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Difference between revisions of "2000 USAMO Problems/Problem 1"

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Let <math>g(a) = \frac{f(a+b) - f(a)}{b}</math>, which is the slope of the secant between <math>(a,f(a))(a+b,f(a+b))</math>. Let <math>b</math> be arbitrarily small; then it follows that <math>g(a+b) - g(a) > 4</math>, <math>g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4</math>. Summing these inequalities yields <math>g(a+kb)-g(a) > 4k</math>. As <math>k \rightarrow \infty</math> (but <math>b << \frac{1}{k}</math>, so <math>bk < \epsilon</math> is still arbitrarily small), we have <math>\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty</math>. This implies that in the vicinity of any <math>a</math>, the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.   
 
Let <math>g(a) = \frac{f(a+b) - f(a)}{b}</math>, which is the slope of the secant between <math>(a,f(a))(a+b,f(a+b))</math>. Let <math>b</math> be arbitrarily small; then it follows that <math>g(a+b) - g(a) > 4</math>, <math>g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4</math>. Summing these inequalities yields <math>g(a+kb)-g(a) > 4k</math>. As <math>k \rightarrow \infty</math> (but <math>b << \frac{1}{k}</math>, so <math>bk < \epsilon</math> is still arbitrarily small), we have <math>\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty</math>. This implies that in the vicinity of any <math>a</math>, the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.   
  
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== See also ==
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== See Also ==
 
{{USAMO newbox|year=2000|before=First question|num-a=2}}
 
{{USAMO newbox|year=2000|before=First question|num-a=2}}
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 07:24, 16 September 2012

Problem

Call a real-valued function $f$ very convex if

\[\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|\]

holds for all real numbers $x$ and $y$. Prove that no very convex function exists.

Solution

Let $y \ge x$, and substitute $a = x, 2b = y-x$. Then a function is very convex if $\frac{f(a) + f(a+2b)}{2} \ge f(a + b) + 2b$, or rearranging,

\[\left[\frac{f(a+2b)-f(a+b)}{b}\right]-\left[\frac{f(a+b)-f(a)}{b}\right] \ge 4\]

Let $g(a) = \frac{f(a+b) - f(a)}{b}$, which is the slope of the secant between $(a,f(a))(a+b,f(a+b))$. Let $b$ be arbitrarily small; then it follows that $g(a+b) - g(a) > 4$, $g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4$. Summing these inequalities yields $g(a+kb)-g(a) > 4k$. As $k \rightarrow \infty$ (but $b << \frac{1}{k}$, so $bk < \epsilon$ is still arbitrarily small), we have $\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty$. This implies that in the vicinity of any $a$, the function becomes vertical, which contradicts the definition of a function. Hence 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
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