Let $$X \sim \mathcal{N}(0, 1)$$ and $$x > 0$$.

$$\mathbb{P}(X > x)$$
$$= \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} \,\mathrm{d}t$$
$$\leq \frac{1}{\sqrt{2\pi}} \int_x^\infty \frac{t}{x} e^{-t^2/2} \,\mathrm{d}t$$
$$= \frac{e^{-x^2/2}}{x \sqrt{2\pi}}.$$

Quite elegant!

Another elegant bound that's stronger for small x is

P(|X| > x) ≤ exp(-x² / 2)

for X ~ N(0, 1) and x >0.

(but I'm not aware of a proof quite as short as the previous one)

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