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Let a(x) = N(0,1) and let b(x) = C(0,1). I would like to prove that the ratio of $$\frac{a(x)}{b(x)} \le \sqrt{\frac{2\pi}{e}}$$

Doing algebraic simplification yield something like $$(1 + x^2)e^{\frac{1}{2} - \frac{x^2}{2}} \le 2$$ I am stuck here and don't know how to proceed further.

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I would use Calculus: take the derivative of $\frac{a(x)}{b(x)}$, find the critical points where the derivative is zero, and determine the maximum. This will imply a global maximum, since the function is continuous on $\mathbb{R}$ and in the limit $$ \lim_{x\rightarrow\pm\infty} \frac{a(x)}{b(x)}=0. $$ Hence, the largest local maximum is your global maximum.

Briefly, \begin{align} \frac{d}{dx}\frac{a(x)}{b(x)} &= \frac{d}{dx} \sqrt{\frac{\pi}{2}}(1+x^2)\mathrm{e}^{-x^2/2}\\ &= \sqrt{\frac{\pi}{2}}x(1-x)(1+x)\mathrm{e}^{-x^2/2}. \end{align} The critical points are $x=-1,0,1$. You can check that at $x=\pm1$, $\frac{a(x)}{b(x)}=\sqrt{\frac{2\pi}{\mathrm{e}}}$.

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