Question: Is there any particular inequality involving $e^{x^2}$ that I need to use here? In any case, I couldn't solve it.

Answer: 
$$\int\_{0}^{x}e^{x^2-t^2}\,dt \stackrel{t=xs}{=} x\int\_{0}^{1}e^{x^2(1-s^2)}\,ds > x\int\_{0}^{1}1\,ds = x.$$
$e^{-t^2}$ is strictly decreasing for positive $t$. So the function you're integrating is always at least $e^{-x^2}$ for $t$ between $0$ and $x$. So $\int\_0^xe^{-t^2}dt > \int\_0^xe^{-x^2}dt = xe^{-x^2} - 0e^{-x^2} = xe^{-x^2}$.