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== Problem ==
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>

holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very [[convex]] [[function]] exists.

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

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

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.  

== See also ==
{{USAMO newbox|year=2000|before=First question|num-a=2}}

[[Category:Olympiad Algebra Problems]]