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? 



An elementary functional inequality
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.


