Integral estimate: upper bound normal distribution I'm looking at the proof of Donsker's theorem and it is used that 
$\frac{1}{\delta}\int_{\frac{1}{\sqrt{\delta}}}^\infty{e^{-y^2/2}}dy\leq \int_{\frac{1}{\sqrt{\delta}}}^\infty{y^2e^{-y^2/2}}dy$ 
But I can't figure out why this should hold. $\delta$ is supposed to tend to $0$. I tried partial integration and change of variable without success. Is there some other fact to use?

Notice that the integral is over $y \geqslant 1/\sqrt{\delta}$. Hence in the integrand, $y^2>\frac{1}{\delta}$, so
$$ \int_{\delta^{-1/2}}^{\infty} y^2 f(y) \, dy \geqslant \int_{\delta^{-1/2}}^{\infty} \frac{1}{\delta} f(y) \, dy. $$
More generally, if $g(x)$ is an nondecreasing function of $x$, you have
$$ g(a) \int_a^b f(x) \, dx \leqslant \int_a^b g(x) f(x) \, dx \leqslant g(b) \int_a^b f(x) \, dx $$
in exactly the same way. This is on the way to the Mean Value Theorem for integrals.
