How can I show that $$\inf_{z>0}\{z^{-1}(z\mu+0.5z^2\sigma^2-\ln(\alpha)\} = \mu+\sigma\sqrt{-2\ln(\alpha)}\quad ?$$
$\alpha \in\ ]0,1]$
Think I'm probably doing something wrong when computing the infimun.
The original full proof I want to do is to show that $\inf_{z>0}\{z^{-1}(\ln(\frac{E[e^{zX}]}{\alpha}))\}= \mu+\sigma\sqrt{-2\ln(\alpha)}$ for $X$ normally distributed $N(\mu,\sigma^2)$
From $AM\geq GM$ we get that for $a,b,z>0$, we have $\frac{1}{2}\left(az+\frac{b}{z}\right)\geq \sqrt{(az)\frac{b}{z}}$, or equivalently, $$az+\frac{b}{z}\geq 2\sqrt{ab}$$ In particular $\mu+0.5\sigma^2 z+\frac{-\ln(\alpha)}{z}\geq \mu+2\sqrt{0.5\sigma^2(-\ln\alpha)}=\mu+\sigma\sqrt{-2\ln\alpha}$.