2000 USAMO Problems/Problem 1

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.

2000 USAMO (Problems • Resources)
Preceded by First question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search

2000 USAMO Problems/Problem 1

Problem

Solution

See also