# 1993 USAMO Problems/Problem 3

## Problem 3

Consider functions $f : [0, 1] \rightarrow \mathbb{R}$ which satisfy

 (i) $f(x)\ge0$ for all $x$ in $[0, 1]$, (ii) $f(1) = 1$, (iii) $f(x) + f(y) \le f(x + y)$ whenever $x$, $y$, and $x + y$ are all in $[0, 1]$.

Find, with proof, the smallest constant $c$ such that $f(x) \le cx$

for every function $f$ satisfying (i)-(iii) and every $x$ in $[0, 1]$.

## Solution

My claim: $c\ge2$

Lemma 1) $f\left(\left(\frac{1}{2}\right)^n\right)\le\left(\frac{1}{2}\right)^n$ for $n\in \mathbb{Z}, n\ge0$

For $n=0$, $f(1)=1$ (ii)

Assume that it is true for $n-1$, then $f\left(\left(\frac{1}{2}\right)^{n}\right)+f\left(\left(\frac{1}{2}\right)^{n}\right)\le f\left(\left(\frac{1}{2}\right)^{n-1}\right)\le \left(\frac{1}{2}\right)^{n-1}$ $f\left(\left(\frac{1}{2}\right)^{n}\right)\le \left(\frac{1}{2}\right)^{n}$

By principle of induction, lemma 1 is proven.

Lemma 2) For any $x$, $\left(\frac{1}{2}\right)^{n+1} and $n\in \mathbb{Z}$, $f(x)\le\left(\frac{1}{2}\right)^n$. $f(x)+f\left(\left(\frac{1}{2}\right)^n-x\right)\le f\left(\left(\frac{1}{2}\right)^{n}\right)\le \left(\frac{1}{2}\right)^{n}$ (lemma 1 and (iii) ) $f(x)\le\left(\frac{1}{2}\right)^n$ (because $f\left(\left(\frac{1}{2}\right)^n-x\right)\ge0$ (i) ) $_\forall 0\le x\le 1$, $\left(\frac{1}{2}\right)^{n-1}\ge2x\ge \left(\frac{1}{2}\right)^n\ge f(x)$. Thus, $c=2$ works.

Let's look at a function $g(x)=\left\{\begin{array}{ll}0&0\le x\le \frac{1}{2};\\1&\frac{1}{2}

It clearly have property (i) and (ii). For $0\le x\le\frac{1}{2}$ and WLOG let $x\le y$, $f(x)+f(y)=0+f(y)\le f(y)$

For $\frac{1}{2}< x\le 1$, $x+y>1$. Thus, property (iii) holds too. Thus $g(x)$ is one of the legit function. $\lim_{x\rightarrow\frac{1}{2}^+} cx \ge \lim_{x\rightarrow\frac{1}{2}^+} g(x)=1$ $\frac{1}{2}c>1$ $c>2$ but approach to $2$ when $x$ is extremely close to $\frac{1}{2}$ from the right side. $\mathbb{Q.E.D}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 