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

Question

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.

Answer 1

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

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

which gives a contradiction.