Find a monotone increasing nonnegative function such that $f'(x)^2 \ge \alpha f(x)f''(x),\alpha>1$

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 $$f(x) for all $$x.

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.

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

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 $$x$$.