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: 
**Hint**. One may differentiate
$$
f(x)=\int\_{0}^{x}e^{-t^2}\,dt-xe^{-x^2} 
$$ obtaining
$$
f'(x)=2x^2e^{-x^2}>0,\quad x>0,
$$ then one may use $f(0)=0$.
Set $e^{-t^2}=u$ and $dt=dv$ we have
$$\int\_{0}^{x}e^{-t^2}dt=xe^{-x^2}+2\int\_{0}^{x}t^2e^{-t^2}dt$$
since $x>0$ we have
$$\int\_{0}^{x}e^{-t^2}dt> xe^{-x^2}$$