Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it Brazil

$f(x)$ is a real-valued function defined on the positive reals such that
(1) if $x < y$ , then $f(x) < f(y)$ ;
(2) $f\left(\frac{2xy}{x+y}\right) = \frac{f(x)+f(y)}{2}$ for all $x$ .
Show that $f(x) < 0$ for some value of $x$ .

Solution

Put $x_n = 1/n$ , $y_n = f(x_n)$ . We have $\frac{2x_{n-1}x_{n+1}}{x_{n-1} + x_{n+1}} = x_n$ , so $y_n = \frac{y_{n-1} + y_{n+1}}{2}$ , or $y_n - y_{n+1} = y_{n-1} - y_n$ . Now $1/2 < 1$ , so $y_2 < y_1$ .

Put $y_1 - y_2 = d > 0$ . Then $y_n - y_{n+1} = d$ for all $n$ . Hence $y_{n+1} = y_1 - nd$ . So $y_n$ is negative for sufficiently large $n$ .

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