# $f(x)^2 ≥ f(x + y)(f(x) + y)$ for no $f$?

Prove that there is no function $f : \mathbb{R}^+ → \mathbb{R}^+$ such that $$f(x)^2 ≥ f(x + y)(f(x) + y)$$ for all $x, y > 0$.

I can't think of a way of solving this.

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Would you mind to gives us a source for this problem? –  Phira May 24 '12 at 14:56

Re-write the inequality a bit and you get

$$1 \geq \frac{f(x+y)}{f(x)} \left( 1 + \frac{y}{f(x)}\right) > \frac{f(x+y)}{f(x)}$$ for any $x,y,f(x), f(x+y) > 0$ which implies that $f(x)$ must be strictly decreasing.

Furthermore by taking $y$ sufficiently large we can find some $x_0 = x + y$ such that $f(x_0) \leq 1$.

Start from such an $x_0$. Consider $y \geq f(x_0)$, we have that $$1 \geq \frac{f(x_0 + y)}{f(x_0)} (1 + 1) \implies \frac{f(x_0 + y)}{f(x_0)} \leq \frac12$$

Hence we have that $$f(x_0+1) \leq \frac12$$

Running the same argument again you get

$$f(x_0 + 1 + \frac12) \leq \frac14$$

and iterating you get that for every $k > 0$ you get

$$f(x_0+2) < f(x_0 + \sum_0^k 2^{-n}) \leq \frac{1}{2^k}$$