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Let $\Phi$ be the cdf and $\phi$ the pdf of the standard normal distribution. I want to show that: $$ \Phi(z)[1-\Phi(z)]\geq \phi(z)^2, \quad z\in\mathbb R. $$ How can I do this? I tried by looking at the derivative, but the second derivative is quite messy and it seems like there should be an easier and nicer way. I've plotted it in Matlab, so I'm certain it should hold, I just want to show why.

In order to avoid unnecessary confusion, the inequality reads: $$ \left[\int_{-\infty}^z\frac{1}{\sqrt{2\pi}}\exp\left\{-t^2/2\right\}dt\right]\left[1-\int_{-\infty}^z\frac{1}{\sqrt{2\pi}}\exp\left\{-t^2/2\right\}dt\right]\geq \frac{1}{2\pi}\exp\{-z^2\}. $$ Hints and ideas are very welcome!

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By Marshall & Olkin (2007) p. 437, Lemma B4, the hazard rate of the standard normal distribution is log-concave. This implies that $$ \frac{\phi \left( s\right) }{1-\Phi \left( s\right) }\frac{\phi \left( -s\right) }{1-\Phi \left( -s\right) }\leq \left( \frac{\phi \left( 0\right) }{1-\Phi \left( 0\right) }\right) ^{2}=\frac{2}{\pi }<1. $$ Then $$ \frac{\phi ^{2}\left( s\right) }{\Phi \left( s\right) \left( 1-\Phi \left( s\right) \right) }=\frac{\phi \left( s\right) }{1-\Phi \left( s\right) }% \frac{\phi \left( -s\right) }{1-\Phi \left( -s\right) }<1 $$ as desired.

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