Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Soviet Union

Problem:

$f$ is a function on the closed interval $[0, 1]$ with non-negative real values. $f(1) = 1$ and $f(x + y) \geq f(x) + f(y)$ for all $x, y$ . Show that $f(x) \leq 2x$ for all $x$ . Is it necessarily true that $f(x) \leq 1.9x$ for all $x$ ?

Solution

Solution:

We have $f(x) = f(1) - f(1 - x) \leq f(1) = 1$ . So for $x \geq 1/2$ , $f(x) \leq 1 \leq 2x$ .

If $x < 1/2$ , then for some $n$ we have $1/2^{n+1} \leq x < 1/2^n$ . Hence by a trivial induction $f(2^n x) \geq 2^n f(x)$ . But $f(2^n x) \leq 1$ , so $f(x) \leq 1/2^n \leq 2x$ .

Note that $f(x) = 0$ for $x \leq 1/2$ and $1$ for $x > 1/2$ satisfies the conditions. But $f(0.51) = 1 > (1.9)(0.51)$ .

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