Consider that for a random variable with any density above $x$, $E(X|X>x) > x$

For a standard normal variate, you should be able to show that $E(X|X>x) = \phi(x)/[1-\Phi(x)]$

Hence $\phi(x)/[1-\Phi(x)]>x$ or $\phi(x)>x[1-\Phi(x)]$ which is stronger than your result.

Consider that for a random variable with any density above $x$, $E(X|X>x) > x$

For a standard normal variate, you should be able to show that $E(X|X>x) = \phi(x)/[1-\Phi(x)]$

(e.g. for the integral in the numerator use the fact that $x\phi(x)=-\phi'(x)$)

Hence $\phi(x)/[1-\Phi(x)]>x$ or $\phi(x)>x[1-\Phi(x)]$ which is stronger than your result.