Upper bound of normal cdf Random variable $X\sim N(0,1)$. Show that, $P(X\geq c) \leq e^{-ct+ \frac{t^{2}}{2}}$ for $c>0$ and for all $t$ in $R$.
I found that $P(X\geq c) = \Phi(-c)$ where $\Phi(x)=\int_{-\infty}^{x}\phi(u)du$ is the cdf of a standard normal variable.
I understand that the upper bound is in the form of mgf of normal distribution. How do I arrive at the above relation?
 A: For any random variable $X$ with moment generating function $M(t)$ existing in an open interval enclosing $t$, say $t\in(-h,h)$, it is true that $$P(X\ge c)\le e^{-ct}M(t)\quad,\,\text{ if }0<t<h$$
This is because 
\begin{align}
M(t)=\int_{\mathbb R} e^{tx}\,dF(x)&\ge \int_c^\infty e^{tx}\,dF(x)
\\&\ge e^{ct}\int_c^\infty \, dF(x)&,\text{ for }0<t<h
\\&=e^{ct}P(X\ge c)
\end{align}
I am not sure if this holds for any $t\in\mathbb R$ in case $X$ is normal.
A: Interesting question. If it is for all $t$, then it must satisfy the worst case, i.e. $t=c$ which minimizes the exponent. Then the upper bound to prove becomes the following: $$P(X\geq c)\leq e^{-c^2/2}$$
From this post, we have the following inequality:
$$P(X\geq c)\leq \frac{e^{-c^2/2}}{\sqrt{2\pi}c}$$
When $c\geq \frac{1}{\sqrt{2\pi}}$, the denominator will be larger than $1$, and RHS will be automatically smaller than $e^{-c^2/2}$. 
When $c\in [0,1/\sqrt{2\pi}]$, $e^{-c^2/2}\in [e^{-1/\pi},1]\approx[0.727,1]$. And, trivially, we know that $P(X\geq c)$ cannot be larger than $0.5$.  
