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

*

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


*$$\color{purple}{b.~~~~~  e ^{-\frac{x^{2}}{2}}\leq\sqrt{e}e ^{-|x|}}$$
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!
 A: Let $f(u)=(1+u)e^{-u/2}$, then $f'(u)=\left(\frac{1}{2}-\frac{u}{2}\right)e^{-u/2}$ thus the maximum of $f$ on $\mathbb{R}^+$ is $f(1)=\frac{2}{\sqrt{e}}$. Thus $\forall x\in\mathbb{R},f(x^2)\leqslant\frac{2}{\sqrt{e}}$, this means that
$$ \forall x\in\mathbb{R},e^{-\frac{x^2}{2}}\leqslant\frac{2}{\sqrt{e}}\frac{1}{1+x^2} $$
A: To show that $$e^{-x^2/2} \leq \frac{2}{\sqrt{e}} \frac{1}{1+x^2}$$ directly via calculus would be painful, since the derivatives of the RHS get annoying.
However, noting that both sides are positive, it is equivalent to show that
$$e^{x^2/2} \geq \frac{\sqrt{e}}{2}(1+x^2). $$
A good strategy for this is to minimize the function
$$f(x) = e^{x^2/2} - \frac{\sqrt{e}}{2}(1+x^2), $$
via the first/second derivative tests, which isn't too bad.
A: 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 $$\frac{x^{2}}{2}-\left|x\right|+\frac{1}{2} = \frac{\left(\left|x\right|-1\right)^{2}}{2}$$
A: Using the "famous" inequality $e^t \geqslant 1+t$,
a) Follows from
$$\sqrt e e^{-\frac{x^2}2} = \frac1{e^{-\frac12+\frac{x^2}2}} \leqslant \frac1{\frac12+\frac{x^2}2}=\frac2{1+x^2}$$
b) Note
$$e^{\frac12+\frac{x^2}2-|x|} \geqslant \frac32+\frac{x^2}2-|x|=\tfrac12(|x|-1)^2+1\geqslant 1$$
A: Let $u = x^2$, and note that the function $f(u) = \sqrt{e^u}$ is convex, so it satisfies $f(u) \geq f(1) + f'(1)(u-1)$, i.e.
$$
\sqrt{e^u} \geq \sqrt{e} + \frac{\sqrt{e}}{2}(u-1) = \frac{\sqrt{e}}{2}(1 + u) \implies e^{-u/2} \leq \frac{2}{\sqrt{e}}\frac{1}{1+u}.
$$
The desired result immediately follows.
