For a large positive integer $n$, define the functions $F,G :[0, 1] \rightarrow [0, 1]$ by $F(t) = (1-t^2)e^{-\frac{n-1}{4}t} + t^2$ and $G(t) = 2(1-t^2)(1-\Phi(\sqrt{n} t)) + t^2$, where $\Phi$ is the standard Gaussian CDF.
Question. What are good estimations for the values of $\min_{0 \le t \le 1}F(t)$ and $\min_{0 \le t \le 1}G(t)$.
I'm hoping for a simple analytic approximation, which would use the fact that $n$ is large.
Here's some back of the envelope calculations that are hopefully okay. For the first one, if $t=o(\ln n/n)$, then the first term is $n^{-o(1)}$, which may tend to $0$, but slower than inverse polynomially in $n$. On the other hand, for $t=\Omega(\ln n/n)$, the second term is clearly $\Omega(\ln^2 n/n^2)$. This suggests that you take $t=8\ln n/n$, which gives $F(t)=\Theta(\ln^2 n/n^2)$. I think this is asymptotically correct by virtue of the above reasoning.
For the second, note that for $x$ large enough, $1-\Phi(x)\sim Ce^{-x^2/2}/x$ for some constant $C>0$ (probably $1/\sqrt{2\pi}$, but let's just write it this way). Here, if we choose $t$ not so small so that $\sqrt{n}t\to \infty$, then $1-\Phi(\sqrt{n}t)=\Theta(\frac{e^{-nt^2/2}}{\sqrt{n}t})$. We may safely assume this, for if $\sqrt{n}t$ remains bounded, then this term is a constant and we can do much better. If we pick $t= C'\sqrt{\ln n/n}$ (which satisfies the condition) for some large enough constant $C'$, this term become $O(1/n\sqrt{\ln n})$, while the $t^2$ term becomes $\Theta(\ln n/n)$. If you pick $t=o(\sqrt{\ln n/n}$), but still such that $\sqrt{n}t\to \infty$, then $e^{-nt^2/2}$ is $n^{-o(1)}$ again, while the denominator is at most $\sqrt{n}$, so the first term is at least $n^{-1/2-o(1)}$. Obviously if you take $t=\Omega(\sqrt{\ln n/n})$, then the $t^2$ term is $\Omega(\ln n/n)$. So it seems to me like for the second part, the minimizer is asymptotically $\Theta(\ln n/n)$ attained for $t=C'\sqrt{\ln n/n}$, where $C'$ is a large enough constant.