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Let $f \in C^1[0,\infty)$ be an increasing function with $f(0)>0$, suppose $\int_0^\infty \frac{1}{f(x)+f'(x)} < \infty$, prove that $\int_0^\infty \frac{1}{f(x)} < \infty$.

I find it weird since the behaviour of $f'$ is random, so I don't know how to control $f$ in terms of $f+f'$.

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    $\begingroup$ You ask to prove it, what is the evidence that this is true? $\endgroup$
    – Fedor Petrov
    Commented Sep 12, 2020 at 11:56
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    $\begingroup$ isn't it just a consequence of $f(x)+f'(x)>f(x)>0$ for all $x>0$? $\endgroup$
    – Carlo Beenakker
    Commented Sep 12, 2020 at 12:13
  • $\begingroup$ @CarloBeenakker no, it seems that it is not $\endgroup$
    – Fedor Petrov
    Commented Sep 12, 2020 at 12:16
  • $\begingroup$ @FedorPetrov It is an exercise $\endgroup$
    – Focus
    Commented Sep 12, 2020 at 12:23
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    $\begingroup$ In general exercises are not welcome here (although I like this one) $\endgroup$
    – Fedor Petrov
    Commented Sep 12, 2020 at 12:58

1 Answer 1

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For $a\geqslant 0,b>0$ we have $$\frac1b\leqslant \frac2{a+b}+\frac{a}{b^2}$$ (if $a\leqslant b$, then $\frac1b\leqslant \frac2{a+b}$; if $a\geqslant b$, then $\frac1b\leqslant \frac{a}{b^2}$). Apply this for $b=f$, $a=f'$ and use that $\int f'/f^2=\int (-1/f)'$ converges.

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