How to prove that if $Z$ is standard normal then $\Pr(|Z| \geq t) \leq C e^{-t^2}$ for all $t \geq 0$ 
Proposition:
  If $Z \sim N(0,1)$, then $$\Pr(|Z| \geq t) \leq C \textrm{exp}(-t^2),$$
  for all $t \geq 0$ and some $C >0 $.

Is this Proposition correct?
If it is, then how I can prove it by using Markov's Inequality? 
If I know $\Pr(|Z| \geq t) \leq \frac{E[e^{t|Z|}]}{e^{t^2}}$.
 A: $$\mathbb{P}(Z>t)=\int_t^\infty(2\pi)^{-1/2}\exp\left(-\frac{1}{2}x^2\right)\,dx$$
$$=\int_t^\infty(2\pi)^{-1/2}\left(\frac{-1}{x}\right)\left(-x\exp\left(-\frac{1}{2}x^2\right)\right)\,dx$$
$$=\left[(2\pi)^{-1/2}\frac{-1}{x} \exp\left(-\frac{1}{2}x^2\right)\right]_t^\infty-\int_t^\infty(2\pi)^{-1/2}\left(\frac{1}{x^2}\right)\left(\exp\left(-\frac{1}{2}x^2\right)\right)\,dx$$
$$=(2\pi)^{-1/2}\frac{1}{t} \exp\left(-\frac{1}{2}t^2\right)-\int_t^\infty(2\pi)^{-1/2}\left(\frac{1}{x^2}\right)\left(\exp\left(-\frac{1}{2}x^2\right)\right)\,dx$$
Now:
$$\int_t^\infty(2\pi)^{-1/2}\left(\frac{1}{x^2}\right)\left(\exp\left(-\frac{1}{2}x^2\right)\right)\,dx\le \int_t^\infty(2\pi)^{-1/2}\left(\frac{1}{t^2}\right)\left(\exp\left(-\frac{1}{2}tx\right)\right)\,dx$$
which can fairly easily be shown to be $=2(2\pi)^{-1/2}t^{-3}\exp\left(-\frac{1}{2}t^2\right)$
Thus,
$$\mathbb{P}(Z>t)= (2\pi)^{-1/2}\frac{1}{t} \exp\left(-\frac{1}{2}t^2\right)\left(1-\frac{2\theta(t)}{t^2}\right)$$
where $0\le \theta(t) \le 1.$
Symmetry gives that:
$$\mathbb{P}(\vert Z \vert>t)= 2(2\pi)^{-1/2}\frac{1}{t} \exp\left(-\frac{1}{2}t^2\right)\left(1-\frac{2\theta(t)}{t^2}\right)\le (2\pi)^{-1/2} \exp\left(-\frac{1}{2}t^2\right)\text{ for }t\ge2$$
Clearly $\mathbb{P}(\vert Z \vert>t)\exp\left(\frac{1}{2}t^2\right)$ is continuous and thus bounded on $0\le t \le2$ by say, $C_1$, thus $\mathbb{P}(\vert Z \vert>t)\le C_1\exp\left(-\frac{1}{2}t^2\right)$ for $0\le t\le 2$.
Taking $C=\max(C_1,(2\pi)^{-1/2}$, we have a uniform bound $\mathbb{P}(\vert Z \vert>t)\le C\exp\left(-\frac{1}{2}t^2\right)$ for all $t$.
Alternatively, Markov gives 
$$\mathbb{P}(Z>t)=\mathbb{P}(e^{uZ}>e^{ut})\le\frac{\mathbb{E} e^{uZ}}{e^{ut}}=e^{\frac{1}{2}u^2-ut}=e^{\frac{1}{2}[(u-t)^2-t^2]}=e^{\frac{-t^2}{2}} \text{ if } u=t$$
