What are the functions for which ${f f''\over f'^2} < 2$? What are the functions $f$ on $[0,1]$ (with continuous first and second derivatives) that satisfy the following conditions:


*

*Monotonically increasing in $[0,1]$, with $f(0)\geq 0$ and $f(1)=1$;

*for all $x\in[0,1]$: 
$$
{f(x) f''(x) \over f'(x)^2} < 2
$$
?
Some simple examples are:


*

*$f(x) = x^k$ for some $k> 0$. Then: $f(x)f''(x)=k(k-1)x^{2k-2}$ and $f'(x)^2 = k^2 x^{2k-2}$ so the quotient is $(k-1)/k < 1 < 2$.

*$f(x) = e^{k (x-1)}$ for some $k\geq 0$. Then: $f(x)f''(x) = k^2 f(x) = f'(x)$ so the quotient is $1 < 2$.
Is there a general form of functions that satisfy these conditions?
 A: Since $f(x), f'(x)>0$ on [0,1]
$$
\frac{f''(x)}{f'(x)}<2\frac{f'(x)}{f(x)}
$$
Then we can write
$$
\frac{f''(x)}{f'(x)}=2\frac{f'(x)}{f(x)}-g(x)
$$
where $g(x)>0$ on [0,1] but is otherwise arbitrary. Integrating from dummy variable t = x to 1:
$$
\ln{|f'(1)|}-\ln{|f'(x)|}=2\ln{|f(1)|}-2\ln{|f(x)|}-G(x)
$$
where $G(x)=\int_x^1{g(t)dt}$ is monotonically decreasing. Since $f(x), f'(x)>0$ on [0,1] the absolute values are irrelevant. Noting that $f(1)=1$, rearranging and taking exponents:
$$
\frac{f'(x)}{f(x)^2}=f'(1)e^{G(x)}
$$
Integrating from x to 1:
$$
\frac{-1}{f(1)}+\frac{1}{f(x)}=f'(1)\int_x^1{e^{G(t)}dt}
$$
or
$$
f(x)=\left[1+f'(1)\int_x^1{e^{G(t)}dt}\right]^{-1}
$$
As an example, choose $g(x)=1$ and $f'(1)=1$. Then
$$
G(x)=1-x
$$
(note that $G$ is monotonically decreasing)
$$
\int_x^1{e^{G(t)}dt}=e^{1-x}-1
$$
$$
f(x)=e^{x-1}
$$
Different choices of $g$ give different results for $f$. Hope that helps.
A: If this helps: (too long for a comment)
$$\left(\frac1f\right)''=-\left(\frac{f'}{f^2}\right)'=-\frac{f''f^2-2ff'^2}{f^4}=\left(2-\frac{f''f}{f'^2}\right)\frac{f'^2}{f^3}.$$
So working with $g:=1/f$ might be easier.
$$gg''=\left(2-\frac{f''f}{f'^2}\right)g'^2$$ is positive.
