Inequality in functional analysis I have a little question about functional analysis, limit and inequality.

Let a function $f$ with domain $]0,+\infty[$ and codomain $]0,+\infty[$ and twice differentiable with the following inequality :
  $$f'+f''\geq f^2$$
Show that the limits of $f$ when $x$ tends to infinity exists and determine it.

My try :

I make the following substition :
  $$f(x)=-g(-e^{-x})$$
  And the inequality becomes :
  $$g(-e^{-x})^2\leq e^{-2x}(-g''(-e^{-x}))$$
  Or 
  $$g(X)^2\leq X^2(-g''(X))$$
  if we put $X=-e^{-x}$

After that, I think we can use a Wirtinger-like type inequality (perhaps generalized) but I don't know how.
If you can read in french see here.
Thanks a lot. 
 A: A partial solution.  If $f(x)$ has a finite limit $L$ as $x\to\infty$, then $L=0$.  For $f'(x)+f''(x) = e^{-x} (d/dx) \,(\exp(x)f'(x)).$  Integrating the inequality $f'+f''\ge f^2$ we get $$e^xf'(x) - e^1 f'(1) \ge \int_1^x e^t f^2(t)\,dt$$ and
$$ f'(x)\ge e^{1-x}f'(1) + \int_1^x e^{t-x} f^2(t)\,dt.$$  If $f$ has limit $L$, the right hand side is, for all $x$ sufficiently large, greater than $L^2/2$.  Hence $f(x)\ge A + (L^2/2) x$ for all $x$ sufficiently large, which is possible only if $L=0$.
A: If $f'(x)\le 0$ for all $x>0$, then $f$ is decreasing and so it has a limit $L\in [0,\infty)$ as $x\to\infty$. As in the proof of kimchi lover, necessarily, $L=0$. 
It remains to study the case in which $f'(x_0)> 0$ for some $x_0>0$. But then, again as in the proof of kimchi lover (with $1$ replaced by $x_0$) we get 
$$f'(x)\ge e^{x_0-x}f'(x_0) + \int_{x_0}^x e^{t-x} f^2(t)\,dt>0,$$ which shows that $f$ is strictly increasing, and so it has a limit $L\in [0,\infty]$ as $x\to\infty$. But since $f(x)>0$ for all $x>0$, necessarily $f(x)\ge f(x_0)>0$ for all $x>x_0$ and so $L\ge f(x_0)>0$. If $L$ were finite then $L=0$ as in the proof of kimchi lover, which is a contradiction. Thus $L=\infty$.
