<s>[INST] How to see $\cos x \leq \exp(-x^2/2)$ on $x \in [0,\pi/2]$? Can anyone help me with the above inequality? I tried looking at the series expansion and I guess the answer indeed lies there, but I fail to see it.
Thanks
 [/INST] Here is another option to prove the inequality by Taylor expansions as the OP suggested. So we are to prove that

$$
\exp(-x^2/2)\ge\cos x,\quad x\in[0,\pi/2].
$$

By doing Taylor expansions at zero with the Lagrange remainders, we can get the estimations
\begin{eqnarray}
\exp(-t)&=&1-t+\frac{t^2}{2!}-\frac{t^3}{3!}+\underbrace{\frac{\exp(-\xi)}{4!}t^4}_{\ge 0}\ge
1-t+\frac{t^2}{2!}-\frac{t^3}{3!},\\
\cos x&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\underbrace{\frac{\cos\xi}{6!}x^6}_{\ge 0}\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}.
\end{eqnarray}
Now we can write with $t=x^2/2$
$$
\exp(-x^2/2)\ge 1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^6}{48}}\ge
1-\frac{x^2}{2}+\frac{x^4}{8}\color{red}{-\frac{x^4}{12}}=
1-\frac{x^2}{2}+\frac{x^4}{24}\ge\cos x,\quad x\in[0,\pi/2]. 
$$
Here at the second step we have used the bound
$$
\color{red}{-\frac{x^6}{48}}\ge\color{red}{-\frac{x^4}{12}}\iff x^4(2+x)(2-x)\ge 0
$$
which is true for $0\le x\le\pi/2<2$.
</s>