Random variable $X\sim N(0,1)$. Show that, $P(X\geq c) \leq e^{-ct+ \frac{t^{2}}{2}}$

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?

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?