$\quad P(\vert X\vert\ge t)\le \exp(\frac{-t^2}{2})$ where $X$ is normal. I would like to prove that $\forall t>0\quad P(\vert X\vert\ge t)\le \exp(\frac{-t^2}{2})$ where $X\sim\mathcal{N}(0,1).$
I use that $P(\vert X\vert\ge t)=P\bigl(\{X\ge t\}\cup P(\{X\le -t \}\bigr)\le P(X\ge t)+P(X\le -t)=\sqrt{\frac{2}{\pi}}\int_{t}^\infty exp(\frac{-u^2}{2})du.$
Not sure how can I continue now.
 A: Let $Q(t) = \int_t^\infty \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \,\mathrm dx$ denote the complementary standard normal distribution function.  Then, we are asked to prove that for $X \sim \mathcal N(0,1)$ for all $t > 0$,
$$P\{|X| \geq t\} \leq e^{-t^2/2}.\tag{1}$$
Now, from symmetry we have that $P\{|X| \geq t\} = P\{X \geq t\} + P\{X \leq -t\} = 2P\{X \geq t\} = 2Q(t)$ and so what we are asked to prove is that for $t > 0$,
$$Q(t) \leq \frac 12 e^{-t^2/2}.\tag{2}$$
Unfortunately, a straightforward application of the Chernoff bound yields the weaker inequality $Q(t) \leq e^{-t^2/2}$ and so a little more work is needed.
Let $X, Y$ denote independent $\mathcal N(0,1)$ random variables so that we have that 
$$P\{|X| \geq t, |Y| \geq t\}= P\{|X| \geq t\} P\{|Y| \geq t\} = 4Q^2(t).
\tag{3}$$ But, the event $\{|X| \geq t, |Y| \geq t\}$ occurs exactly when the random point $(X,Y)$ lies in one of the four quadrants with corners at $(\pm t, \pm t)$, that is, outside the circle of radius $\sqrt{2} t$ centered at the origin.. Thus we have that
\begin{align}
P\{|X| \geq t, |Y| \geq t\} &\leq P\{X^2+Y^2 \geq 2t^2\}\\
&= \int_{r=\sqrt{2}t}^\infty \int_0^{2\pi}\frac{1}{2\pi}e^{-r^2/2}r \,\mathrm  d\theta \, \mathrm dr\\
&= \int_{r=\sqrt{2}t}^\infty re^{-r^2/2} \, \mathrm dr\\
&= - e^{-r^2/2}\biggr|_{\sqrt{2}t}^\infty\\
&= e^{-t^2}
\end{align}
and so upon substituting into $(3)$, we get that
$$4Q^2(t) \leq e^{-t^2} \implies Q(t) \leq \frac 12 e^{-t^2/2}
\implies P\{|X| \geq t\} \leq e^{-t^2/2}$$
