How to derive this inequality

Question

I learnt that for a standard normal random variable $Z$ and positive $x$, we have $$\mathbb P (Z > x) \geq \frac{x}{x^2+1} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} = \frac{1}{x+\frac{1}{x}} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}. \tag1$$ Then I also read that if $x$ is large enough, we can write the above inequality as $$\mathbb P (Z > x) \geq \frac{c}{x} e^{-\frac{x^2}{2}}, \tag2$$ for some constant $c$. It is clear that $c$ absorbs the $\frac{1}{\sqrt{2\pi}}$. I do not see how to get from $(1)$ to $(2)$.

Answer 1

Re gaussian tails, the question refers to two distinct inequalities, namely, $$\frac{x}{x^2+1}g(x)\lt P(Z\gt x)\lt\frac1{x}g(x),$$ where $g$ is the standard gaussian density, defined by $$g(x)=\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}.$$ The upper bound is easy, note that, for every positive $x$, $$P(Z\gt x)=\int_x^\infty g(t)\mathrm dt\lt\int_x^\infty\frac{t}x g(t)\mathrm dt=\frac1x\left.(-g(t))\right|_x^\infty=\frac1xg(x).$$ That the lower bound implies that, for $x$ large enough, $$P(Z\gt x)\gt\frac{c}{x}g(x),$$ for some positive $c$, is shown as follows: for every $x\gt1$, $\frac1x\lt x$ hence $x+\frac1x\lt2x$, which implies $$P(Z\gt x)\gt\frac1{x+\frac1x}g(x)\gt\frac1{2x}g(x).$$