Consider functions $\, f: [0,1] \rightarrow \mathbb{R} \,$ which satisfy (i) $f(x) \geq 0 \,$ for all $\, x \,$ in $\, [0,1],$ (ii) $f(1) = 1,$ (iii) $f(x) + f(y) \leq f(x+y)\,$ whenever $\, x, \, y, \,$ and $\, x + y \,$ are all in $\, [0,1]$. Find, with proof, the smallest constant $\, c \,$ such that \[ f(x) \leq cx \]for every function $\, f \,$ satisfying (i)-(iii) and every $\, x \,$ in $\, [0,1]$.
Problem
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Tags: function, AMC, USA(J)MO, USAMO, induction, algebra unsolved, algebra
jhaussmann5
09.07.2005 16:07
It's USAMO 1993/3: http://www.kalva.demon.co.uk/usa/usa93.html
aznlord1337
17.02.2009 20:09
This is basically the kalva solution if I remember correctly but the link is broken: Letting $ x = 1-y$ we have $ f(x) + f(1-x) \leq 1$ which means $ f(x) \leq 1$. Let $ x=y \leq 1/2$. Then $ 2f(x) \leq f(2x)$. $ 4f(x) \leq 2f(2x) \leq f(4x)$, induction like this yields $ 2^nf(x) \leq f(2^nx)$ with $ x$ in an appropriate range. If $ \frac{1}{2^{n+1}} < x \leq \frac{1}{2^n}$ for $ 1 \leq n$, then $ 2^nf(x) \leq f(2^nx) \leq 1$. So as $ x$ approaches $ \frac{1}{2^{n+1}}$, $ c$ approaches $ 2$. If $ \frac{1}{2} < x$, then $ f(x) \leq 1$ and as $ x$ approaches $ \frac{1}{2}$, $ \frac{1}{x} = c$ approaches $ 2$. This means $ c=2$ is the minimum. It remains to find a function that fits $ c=2$ which is not very hard.
Aryth
12.04.2009 18:32
aznlord1337 wrote: It remains to find a function that fits $ c = 2$ which is not very hard. $ f(x)= \begin{cases} 0 & 0 \le x \le \frac{1}{2} , \\ 1 & \frac{1}{2} < x \le 1. \end{cases}$