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Let $g$ be a $C^1$ function with $g(0)=0$ and $g(t)>0$ for all $t>0$. I am surprised that for all such $g$ the following seems to hold

$\frac{\int_0^t(g'(s))^2ds}{g^2(t)}\geq \frac{1}{t}$ for all t>0.

I tried to find a counterexample but I was not successful. Is the above true or one can find a counterexample?

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    $\begingroup$ $(tg'-g)^2\ge0$ $\endgroup$
    – Alex Degtyarev
    Commented Jun 26, 2015 at 7:38
  • $\begingroup$ @AlexDegtyarev, do you mean $(t g'(s) - g(t))^2 \ge 0$? $\endgroup$
    – LSpice
    Commented Jun 26, 2015 at 7:44
  • $\begingroup$ No, I think everything is at $t$. Just multiply everything by $g^2$ and differentiate, observing that the inequality holds at $t=0$. $\endgroup$
    – Alex Degtyarev
    Commented Jun 26, 2015 at 8:47

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Well since $g(0) = 0$, we may rewrite the desired inequality as:

$$\left(\int_0^{t} 1\,ds\right)\left(\int_0^{t} (g'(s))^2\,ds\right) \ge \left(\int_0^t g'(s)\,ds\right)^2$$

which is just the Cauchy-Schwarz Inequality.

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