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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.

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  • $\begingroup$ Please define the term "incremental". Also, please show any work you've done. $\endgroup$ – Adrian Keister Jun 19 '19 at 15:09
  • $\begingroup$ The word you want is "increasing", not "incremental". $\endgroup$ – Robert Israel Jun 19 '19 at 15:30
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    $\begingroup$ Oh I have changed it. Thank you very much! $\endgroup$ – zbh2047 Jun 19 '19 at 15:34
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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 $x$.

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