Prove $e ^{-\frac{x^{2}}{2}}\leq \frac{2}{\sqrt{{e}}}\frac{1}{1+x^{2}}$ I have two inequalities today :

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*$$\color{red}{a.~~~~~ e ^{-\frac{x^{2}}{2}}\leq \frac{2}{\sqrt{{e}}}\frac{1}{1+x^{2}}}$$


*$$\color{purple}{b.~~~~~  e ^{-\frac{x^{2}}{2}}\leq\sqrt{e}e ^{-|x|}}$$
I tried to relate these inequalities to this famous inequality
$$e^{x}\geq 1+x$$
but without success so I considere the function :
$$f(x)=e ^{-\frac{x^{2}}{2}}-\frac{2}{\sqrt{e}}\frac{1}{1+x^{2}}$$
But the calculations were too much.

Thanks!

For $a$, you can instead consider $y = x^2$ for positive $y$ to get $$e^{-\frac{y}{2}} \le \frac{2}{\sqrt{e}} \frac{1}{1+y} \to (1+y)e^{-\frac{y}{2}} \le \frac{2}{\sqrt{e}}$$
You can then find the maximum of the left hand side by differentiating and finding the roots. It would end up being at $y = 1$, and $(1+1) e^{-\frac{1}{2}} = \frac{2}{\sqrt{e}} \le \frac{2}{\sqrt{e}}$.
For $b$, you can multiply by $e^{\frac{x^2}{2}}$ on both sides to get $$\sqrt{e}e^{-\left|x\right|}e^{\frac{x^{2}}{2}} = e^{\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2}} \ge 1$$
Taking logs, $$\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2} \ge 0$$
This is easy to show since $$\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2} = \frac{\left(\left|x\right|-1\right)^{2}}{2}$$
