$X \sim \mathcal{N}(0,1)$, then to show that for $x > 0$,
$$
\mathbb{P}(X>x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \>.
$$
Since for $t \geq x > 0$ we have that $1 \leq \frac{t}{x}$,
$$
\mathbb{P}(X > x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty 1 \cdot e^{-t^2/2} \,\mathrm{d}t \leq \frac{1}{\sqrt{2\pi}} \int_x^\infty \frac{t}{x} e^{-t^2/2} \,\mathrm{d}t = \frac{e^{-x^2/2}}{x \sqrt{2\pi}} . 
$$