TITLE: Find a monotone increasing nonnegative function such that $f'(x)^2 \ge \alpha f(x)f''(x),\alpha>1$
QUESTION [6 upvotes]: Can we construct a function that is monotone increasing and nonnegative, such that  $f'(x)^2 \ge \alpha f(x)f''(x)$ for each $x\in \mathbb R$, where $\alpha$ is greater than $1$. If not, how can we give a proof?
Note: we say $f(x)$ is monotone increasing, iff $f(x)<f(y)$ for all $x<y$.
I have tried a lot of examples but havn't found a solution. For example, consider $f(x)=b\exp(ax)$, then $f'(x)^2=f(x)f''(x)=b^2a^2\exp(ax)$, so the constraint "$\alpha$ is greater than 1" is not true.

REPLY [6 votes]: If $f > 0$, we can write $f(x) = \exp(g(x))$ where $g$ is monotone increasing.  The inequality then becomes $$ \alpha ((g')^2 + g'') \le (g')^2 $$
i.e. with $g' = u$,
$$ u' \le (1/\alpha - 1) u^2 $$
Note that $1/\alpha - 1 < 0$, so $u$ is always positive and $u'$ is always negative.  But
$$\dfrac{d}{dx} \dfrac{1}{u} = - \frac{u'}{u^2} \ge 1 - \frac{1}{\alpha}  > 0$$
and hence for $x < 0$,
$$ \frac{1}{u(x)} \le \frac{1}{u(0)} + \left(\frac{1}{\alpha}-1\right) x$$
which means we will reach $1/u(x) = 0$ at a finite value of $x$, i.e. the solution can't exist for all real $x$.