Find $f$ such that $(f(x))^2 \ge f(x + y)(f(x) + y), \forall x, y$

Question

Find all $f : (0, \infty) \rightarrow (0, \infty)$ such that $(f(x))^2 \ge f(x + y)(f(x) + y), \forall x, y \gt 0$

---

My guess there is no such function but I cannot prove it.

The most obvious idea is to make $y=x$ but it seem to lead nowhere.

Also, it's easy to show $f$ is strictly decreasing

Answer 1

Suppose that such a function $f$ exists. It is easily seen that $f$ is strictly decreasing. Let $a_0:=1$ and, for $k=1,2,\ldots$, write $$a_k:=a_{k-1}+f\left(a_{k-1}\right)\,.$$ From the given inequality, plugging in $x:=a_{k-1}$ and $y:=f\left(a_{k-1}\right)$ yields $$\big(f\left(a_{k-1}\right)\big)^2\geq f\left(a_{k}\right)\,\big(2\,f\left(a_{k-1}\right)\big)\,,$$ whence $$f\left(a_k\right)\leq \frac{1}{2}\,f\left(a_{k-1}\right)$$ for all $k=1,2,\ldots$. By induction on $k=0,1,2,\ldots$, $$f\left(a_k\right)\leq \frac{b}{2^k}\,,$$ where $b:=f(1)$. Now, note that $$a_k-a_{k-1}=f\left(a_{k-1}\right)\leq \frac{b}{2^{k-1}}$$ for each $k=1,2,\ldots$. We have $$a_k= a_0+ \sum_{j=1}^k\,\left(a_j-a_{j-1}\right)< 1+\sum_{j=1}^\infty\,\frac{b}{2^{j-1}}=1+2b\,.$$ This shows that $$f(1+2b)<f\left(a_k\right)\leq \frac{b}{2^k}$$ for all $k=0,1,2,\ldots$. However, as $k\to\infty$, the upper bound goes to $0$, so that $$f(1+2b)\leq 0\,.$$ However, this contradicts the requirement that $f(1+2b)>0$. Ergo, there is no function $f$ with the required property.

Answer 2

Rearranging since $x, y, f > 0$, we have

$\frac{f\left(x+y\right)-f\left(x\right)}{y}\leq\frac{-f\left(x+y\right)}{f\left(x\right)}$

take the limit (if it exists) as $y$ goes to $0$ and $f'(x)\leq-1$. The function is strictly decreasing (i.e. even where $f'$ doesn't exist) so if $f(1)=a$, then $f(1+a)\leq0$ which is a contradiction.