Upper bound inequality cumulative normal distribution

Question

According to this post, I found for $X \sim N(0,1)$, $x > 0$ the result that

\begin{align} \frac{1}{\sqrt{2\pi}}\big(\frac{1}{x}-\frac{1}{x^3}\big)e^{-\frac{x^2}{2}} \leq P(X>x) \leq \frac{1}{\sqrt{2\pi}}\frac{1}{x}e^{-\frac{x^2}{2}}. \end{align}

Now, an even stronger upper bound has to hold, namely \begin{align} P(X>x) \leq e^{-\frac{x^2}{2}}. \end{align} For $x \geq 1$ this is obvious. However, it seems to be rather difficult to show that this inequality does hold for $0<x<1$. What do you think?

Answer 1

$P(X > x) = P(e^{\lambda X} > e^{\lambda x})$ for all $\lambda$.

Now by Markov's inequality, \begin{align} P(e^{\lambda X} > e^{\lambda x}) &\leq \frac{\mathbb{E}[e^{\lambda X}]}{e^{\lambda x}} \\ &=e^{\frac{\lambda^2}{2} - \lambda x}. \end{align} So for $\lambda = x$, \begin{align} P(X > x) \leq e^{\frac{x^2}{2}-x^2} = e^{-\frac{x^2}{2}}. \end{align}