If $f'+f''\geq f^2$ show that $\frac{f'}{f''}\leq 1$ for all $x\in(0,+\infty)$.

I have a question about this :

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 we have $\dfrac{f'}{f''}\leq 1$ for all $x\in$ $]0,+\infty[$

I have no idea to prove or disprove this .

Thanks.

Solution 1:

Make the substitution :

$f(x)=-y(-e^{-x})$ the initial condition becomes with $x \in ]0;\infty[$ :

$y''(-e^{-x})(-e^{-2x})\geq y(-e^{-x})^2$

So the function $y(-e^{-x})$ is concave and now suppose that $y$ is increasing .

We put $t=-e^{-x}$

We can apply a modified version of the Taylor-Lagrange by Wang (here leaf 6) with $a=0$ ,$n=1$ , $0<c<1$ and $-f(x)=y(t)$ we find :

$$y''(c)t^2=(y(t)-y(0)-y'(0)t)2!$$

Moreover for $x=c$ we have :

$$y''(c)c^2=(y(c)-y(0)-y'(0)c)2!$$

But on one hand the function $y$ is concave and negative so :

$y(c)-y(0)-y'(0)c=c(\frac{y(c)-y(0)}{c-0}-y'(0))>0$

On the other hand we have $y''(-e^{-x})<0$ so it's a contradiction and the function is decreasing.

So we have $\frac{f'}{f''}\leq0<1$.