Prove function (0, ∞) to (0, ∞) can't exist if $(f(x))^2>(f(x+y))((f(x)+y))$ So we're trying to prove that no function exists $f: (0,\infty) \rightarrow(0,\infty)$ such that $f(x)^2\ge f(x+y) (f(x)+y)\quad x,y\gt0$
I've tried to break it up into 3 cases to show if $x\gt y$ and  $x\lt y$ and $x= y$ to show that $f(x+y)\gt f(x)$, but doesn't that depend on the initial function? i.e $\frac{1}{x}\gt\frac{1}{x+y}$ not $\lt$ like I wanted to show. 
I'm just stumped at how to prove this for any function. 
We've tried to set it all up on one side and do the quadratic equation to try and chase down a numerical value to prove that $f(x+y)\gt f(x)$ so we could then say that $f(x+y)f(x)\gt f(x)f(x)$ which would imply that $f(x+y)f(x)\gt f(x)^2$ so our original function of $f(x)^2\ge((f(x+y))((f(x)+y))$ could be rewritten as $f(x)^2\ge((f(x))^2+f(x+y)y)$ which we know $f(x)^2 \lt f(x)^2 + y$ so we'd have a contradiction.
But I can never seem to figure out how to write a formal proof of the ideas I have. If my idea was even a valid idea. Thanks ahead of time!
Just had an idea to find some value such that it is always contradictory like choosing $y=x^2-x$, but then $y$ could have a negative value so I'm going to have to find a new y=, but that's the path I'm going now!
 A: The idea behind the proof is as follows. $f(x)$ is decreasing positive function with the limit $0$ at infinity. We try to show that with the inequality stated above, $f(x)$ decreases so fast that it cannot stay all the time above zero.

First observe that $f$ is monotonically decreasing:
$$
(f(x))^2\geq f(x+y)((f(x)+y))\geq f(x+y)f(x)\implies f(x)\geq f(x+y)
$$
Secondly observe that:
$$
(f(x))^2\geq f(x+y)((f(x)+y))\geq f(x+y)y\implies f(x+y)\leq \frac{f(x)^2}{y}
$$
Therefore $\displaystyle\lim_{x\to+\infty}f(x)=0$.
Now because $f$ is monotonically decreasing, it has countable point of discontinuities. Suppose for the rest that the values of $x,y$ are chosen in the set of continuities of $f$, denoted by $\mathcal C$. $f$ is over $\mathcal C$ differentiable. We have:
$$
\frac{f(x+y)-f(x)}{y}+\frac{f(x+y)}{f(x)}\leq 0
$$
When $y\to 0$ for $x\in\mathcal C$, we have:
$$
f'(x)+1\leq 0.
$$
Therefore $g(x)=f(x)+x$ is decreasing on $\mathcal C$ and because $f(x)$ and $x$ are positive, therefore $g(x)\geq 0$, bounded from below and hence it should converge to $L<\infty$ when $x\to\infty$. But:
$$
\displaystyle\lim_{x\to+\infty}g(x)=\displaystyle\lim_{x\to+\infty}f(x)+\displaystyle\lim_{x\to+\infty}x=0+\infty=\infty
$$
which is a contradiction.
