Can it be proved that $\frac{d}{d x} \frac{( 1 - \Phi(x)) }{\phi(x)} < 0$? (updated question with more contents)

Question

Let $\phi(x) = \frac 1{\sqrt{2\pi}}e^{- \frac 12 x^2}$ and $\Phi(x)=\int_{-\infty}^x \phi(t)dt$ be the PDF and CDF of the standard normal distribution.

Can it be proved that $$ \frac{d}{d x} \frac{( 1 - \Phi(x)) }{\phi(x)} < 0 \ ?$$

To show that i find $$ \frac{d}{d x} \frac{( 1 - \Phi(x)) }{\phi(x)} = \frac{-\phi(x)^2 -[-x \phi(x) ( 1 - \Phi(x)) ]}{\phi(x)^2} \ $$

which is

$$ \frac{d}{d x} \frac{( 1 - \Phi(x)) }{\phi(x)} = \frac{-\phi(x) +[x ( 1 - \Phi(x)) ]}{\phi(x)} \ $$ and $$ \frac{d}{d x} \frac{( 1 - \Phi(x)) }{\phi(x)} = \frac{x ( 1 - \Phi(x)) }{\phi(x)} -1 \ $$

So we need to prove

$$\frac{x ( 1 - \Phi(x)) }{\phi(x)} -1 < 0 \ $$

which is

$$\ 1 - \Phi(x) < \frac{1 }{x} \phi(x) \ $$

And by following @kavi-rama-murthy proof at the accepted answer below, it is proved. Thanks @kavi-rama-murthy

Answer 1

$\int_x ^{\infty} e^{-t^{2}/2} \, dt =\int_x ^{\infty} t \frac 1 t e^{-t^{2}/2} \, dt <\frac 1 x \int_x ^{\infty} t e^{-t^{2}/2} \, dt = \frac 1 x e^{-x^{2}/2}$ so $1-\Phi (x) < \frac 1 x \phi (x)$ This proves teh inequality you are looking for.

Answer 2

You can see this hence the graph