EDIT: Hm. I've just realized that this "answer" does not answer the question at all. This is an application of Grothendieck Duality, not of GRR. I suppose I was not reading the question carefully when I wrote this. Sigh. Sorry. Given the number of up-votes, it seems reasonable to leave it here, but I apologize for the possible confusion I may have caused with it.
Here is a simple proof of Kempf's criterion for rational singularities:
Theorem. (Kempf) Let $X$ be a normal variety over $\mathbb C$. Then $X$ has rational singularities (i.e., for a resolution $\phi:\widetilde X\to X$, $R^i\phi_*\mathcal O_{\widetilde X}=0$ for $i>0$) if and only if $X$ is Cohen-Macaulay and $\phi_*\omega_{\widetilde X}\simeq \omega_X$.
Proof. ($RHom$ stands for sheaf-RHom, $\omega^\cdot$ for the dualizing complex, $n=\dim X$).
First assume that $X$ has rational singularities. Then \begin{multline} \omega_X^\cdot\simeq RHom_X(\mathcal O_X,\omega_X^\cdot)\simeq RHom_X(R\phi_*\mathcal O_{\widetilde X},\omega_X^\cdot) \simeq_{\text{by Grothendieck duality}}\\\ \simeq R\phi_*RHom(\mathcal O_{\widetilde X}, \omega_{\widetilde X}^\cdot) \simeq R\phi_*\omega_{\widetilde X}[n]\simeq \phi_*\omega_{\widetilde X}[n] \end{multline} The last isomorphism follows by Grauert-Riemenschneider vanishing. The two ends of the displayed isomorphism shows that $\omega_X^\cdot$ has only one non-zero cohomology sheaf and hence $X$ is Cohen-Macaulay and that non-zero cohomology sheaf is $\phi_*\omega_{\widetilde X}\simeq \omega_X$.
The other direction is similar: \begin{multline} \mathcal O_X\simeq RHom_X(\omega_X^\cdot,\omega_X^\cdot)\simeq RHom_X(R\phi_*\omega_{\widetilde X}^\cdot,\omega_X^\cdot) \simeq_{\text{by Grothendieck duality}}\\\ \simeq R\phi_*RHom(\omega_{\widetilde X}^\cdot, \omega_{\widetilde X}^\cdot) \simeq R\phi_*\mathcal O_{\widetilde X} \end{multline}
Edit: A little bit more use of G̶R̶R̶ Grothendieck Duality(!) gives you interesting theorems about singularities. See for instance herehere.
