Proof of upper-tail inequality for standard normal distribution

Question

$X \sim \mathcal{N}(0,1)$, then to show that for $x > 0$, $$ \mathbb{P}(X>x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \>. $$

Answer 1

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

Answer 2

Integrating by parts, $$\begin{align*} Q(x) &= \int_x^{\infty} \phi(t)\mathrm dt = \int_x^{\infty} \frac{1}{\sqrt{2\pi}}\exp(-t^2/2) \mathrm dt\\ &= \int_x^{\infty} \frac{1}{t} \frac{1}{\sqrt{2\pi}}t\cdot\exp(-t^2/2) \mathrm dt\\ &= - \frac{1}{t}\frac{1}{\sqrt{2\pi}}\exp(-t^2/2)\biggr\vert_x^\infty - \int_x^{\infty} \left( - \frac{1}{t^2} \right ) \left ( - \frac{1}{\sqrt{2\pi}} \exp(-t^2/2) \right )\mathrm dt\\ &= \frac{\phi(x)}{x} - \int_x^{\infty} \frac{\phi(t)}{t^2} \mathrm dt. \end{align*} $$ The integral on the last line above has a positive integrand and so must have positive value. Therefore we have that $$ Q(x) < \frac{\phi(x)}{x} = \frac{\exp(-x^2/2)}{x\sqrt{2\pi}}~~ \text{for}~~ x > 0. $$ This argument is more complicated than @cardinal's elegant proof of the same result. However, note that by repeating the above trick of integrating by parts and the argument about the value of an integral with positive integrand, we get that $$ Q(x) > \phi(x) \left (\frac{1}{x} - \frac{1}{x^3}\right ) = \frac{\exp(-x^2/2)}{\sqrt{2\pi}}\left (\frac{1}{x} - \frac{1}{x^3}\right )~~ \text{for}~~ x > 0. $$ In fact, for large values of $x$, a sequence of increasingly tighter upper and lower bounds can be developed via this argument. Unfortunately all the bounds diverge to $\pm \infty$ as $x \to 0$.