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Algebra Difficulty 4.8 AIME Prove it Turkey

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(x + f(x)) = f(-x)$
for all real numbers $x$ and $f(x) \le f(y)$ for all real numbers $x \le y$ .

Solution

All constant functions.

Let $a < b$ be two arbitrary numbers. Consider a sufficiently large $x$ so that $x > -a$ and $x > b - f(-a)$ , thus $-x < a$ and $x + f(x) > b - f(-a) + f(-a) = b$ (here we used $f(x) \ge f(-a)$ since $x > -a$ ). Now $-x < a < b < x + f(x)$ while $f(-x) = f(x+f(x))$ , hence $f(a) = f(b)$ , so $f$ is constant.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.