When $Z \sim N(0,1)$, then $P(\vert Z\vert \gt t)\le\sqrt\frac{2}{\pi}\times \frac{exp(-t^2/2)}t$.
My question is: I can get the aboving inequality, but how can I get the inequality such that $P(\vert Z\vert \gt t)\le\sqrt\frac{2}{\pi}\times \frac{t}{1+t^2} \times exp(-t^2/2)$?
We have $$1-\text{erf}(x)=\frac{x}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x^2+(1/2)/q}$$ with $q\geq 1$. Thus $$1-\text{erf}(x)\geq\frac{x}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x^2+(1/2)}$$ Now $$P(|Z|>t)=P(Z>t)+P(Z<-t)\\ =2P(Z>t)\\ =2\left(1-\frac{1}{2}(1+\text{erf}(t/\sqrt{2}))\right)\\ =1-\text{erf}(t/\sqrt{2})$$ $$1-\text{erf}(t/\sqrt{2})\geq \sqrt{\frac{2}{\pi}}\exp(-t^2/2)\frac{t}{t^2+1}$$ This looks right.