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