Upper bound for density of standard Brownian Motion

Question

I need to prove the following: if $\{W_t\}_{t \geq 0}$ is standard Brownian motion under the measure $\mathbb{P}$, then for $x>0$, \begin{align} \mathbb{P}[W_t \geq x]=\int_x^{\infty}\frac{1}{\sqrt{2\pi t}}e^{-y^2/2t} \,dy \leq \frac{\sqrt{t}}{x\sqrt{2\pi}}e^{-x^2/2t}. \end{align} I think I need to use integration by parts, but I don't really know which functions to pick. Does anyone know how to proceed or how to prove this otherwise?

Answer 1

$\int_x^{\infty} e^{-y^{2}/2t} dy=\int_x^{\infty} \frac 1 y ye^{-y^{2}/2t} dy \leq \frac 1 x \int_x^{\infty} ye^{-y^{2}/2t} dy $. Now $\int_x^{\infty} ye^{-y^{2}/2t} dy =-t (e^{-y^{2}/2t})|_x^{\infty}=te^{-x^{2}/2t}$. Divide by a suitable constant to finish.