Verifying the bound $\cos^2x\leq 2 e^{-x^2/4}-1$ A CAS plot suggests that for $x\in[0,\pi/2]$:
$$\cos^2x\leq 2\, e^{-x^2/4}-1.$$
The difference is not increasing in this range so I don't think differentiation alone will do.
Any suggestions welcome.
Note $\cos x\leq e^{-x^2/2}$ in this range.
 A: The given inequality is a consequence of 
$$ f(x)=\frac{\sin^2 x}{x^2} > \frac{2-2e^{-x^2/4}}{x^2}=g(x) \tag{1}$$
for any $x\in\left(0,\frac{\pi}{2}\right)$. $(1)$ is a pretty loose inequality, and both $f(x)$ and $g(x)$ are decreasing and concave functions over $I=\left(0,\frac{\pi}{2}\right)$. In particular
$$ \forall x\in I,\quad f(x)\geq 1-\frac{2}{\pi}\left(1-\frac{4}{\pi^2}\right)x>\begin{smallmatrix}\text{the equation of the tangent line}\\\text{to the graph of }g(x)\text{ at }x=\pi/2\end{smallmatrix}\geq g(x)\tag{2} $$
and $(1)$ is proved.
A: Hint. Split the interval $[0,\pi/2]$ in two halves.
1) Show that for $x\in [0,\pi/4]$,
$$\cos^2(x)\leq 1-\frac{x^2}{2}\leq 2 e^{-x^2/4}-1.$$
Note that the Taylor expansions at $x=0$ of the two functions are
$$\cos^2(x)=1-\frac{x^2}{2} +O(x^4)\quad\mbox{and}\quad 2 e^{-x^2/4}-1=1-x^2+O(x^4).$$
2) Show that for $x\in [\pi/4,\pi/2]$,
$$\cos^2(x)\leq 1-\frac{2x}{\pi}\leq 2 e^{-x^2/4}-1.$$
Note that $\cos^2(x)$ is convex and $2 e^{-x^2/4}-1$ is concave in  such interval.
A: We can actually prove that

$$
2e^{-x^2/4}-1\ge\cos x,\quad\forall x\in[0,\pi/2].\tag{1}
$$

It is a tighter inequality because $\cos x\ge\cos^2x$ in the interval. We have
$$
(1)\iff e^{-x^2/4}\ge \frac{1+\cos x}{2}=\cos^2(x/2)\quad\forall x\in[0,\pi/2].\tag{2}
$$
Now denote $t=x/2\in[0,\pi/4]$ and continue
$$
(2)\iff e^{-t^2}\ge\cos^2t\iff e^{-t^2/2}\ge\cos t,\quad\forall t\in[0,\pi/4].\tag{3}
$$
The last inequality in (3) can be estimated from the Taylor expansions using at the second step that $|t|<1$, hence, $t^6\le t^4$
$$
e^{-t^2/2}\ge 1-\frac{t^2}{2}+\frac{t^4}{8}-\frac{t^6}{48}\ge
1-\frac{t^2}{2}+t^4\Big(\underbrace{\frac{1}{8}-\frac{1}{48}}_{=\frac{5}{48}>\frac{1}{24}}\Big)\ge
1-\frac{t^2}{2}+\frac{t^4}{24}\ge\cos t.
$$

P.S. Explanations of inequalities from Taylor expansions with Lagrange remainders:
\begin{eqnarray}
e^{-x}&=&1-x+\frac{x^2}{2}-\frac{x^3}{3!}+\frac{e^{-\xi}}{4!}x^4\ge1-x+\frac{x^2}{2}-\frac{x^3}{3!},\\
\cos x&=&1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{\cos\xi}{6!}x^6\le1-\frac{x^2}{2}+\frac{x^4}{4!}.
\end{eqnarray}

EDIT: Thanks to Marty Cohen sharing the link to a really elementary proof of (3) by differentiation.
