Show that $\cos^2\left(x\right)\left(2+\cos 2x\right)-2 \cos^2\left(x+\frac{\sin(2x)}{4}\right)\ge0$ Show that for all $x\in\mathbb{R},$
\begin{equation}
\cos^2\left(x\right)\left(2+\cos 2x\right)-2 \cos^2\left(x+\frac{\sin 2x}{4}\right)\ge0
\end{equation}
I have checked this numerically and this is correct but I cannot show analytically. I have used the following useful formulas but had no luck:
\begin{gather}
\sin(2x)=2\sin (x)\cos (x),\newline
\cos (2x)=2\cos^2(x)-1,\newline
\sin^2(x)=1-\cos^2(x).
\end{gather}
I guess there is some bounding trick that I am missing. Any help is appreciated
 A: If you expand the first term on the LHS, and add the second term on the LHS to both sides, you'll get
$2\cos^2(x) + \cos^2(x)\cos(2x) \geq 2\cos^2(x+\frac{\sin(2x)}{4})$
Now, using the identity $\cos(2x)+1 = 2\cos^2(x)$, you get:
$\cos(2x)+1 + \frac{\cos(2x)+1}{2}\cos(2x) \geq \cos(2x+\frac{\sin(2x)}{2})+1$
Subtract $1$ from both sides, and write the LHS by taking the expression into the paranthesis of $\cos(2x)$, you get:
$\cos(2x)(\frac{\cos(2x)+3}{2}) \geq \cos(2x + \frac{\sin(2x)}{2})$
Substitute $y=2x$ at this point, and it is easy to see that for each $y \in \mathbb{R}$ that satisfies the equation, $y + 2\pi k$ and $-y$ also satisfies the equation for $k \in \mathbb{Z}$. Thus, it is enough to consider the inequality for $y \in [0,\pi]$.
$\frac{\cos(y)+3}{2} \geq 1$ for all $y$; thus, if $\cos(y) \geq \cos(y + \frac{\sin(y)}{2})$ for all $y$, the inequality holds.
It can be seen that the equality $\cos(y) = \cos(y + \frac{\sin(y)}{2})$ holds when either $\frac{\sin(y)}{2}=0$ or $2\pi- y = y + \frac{\sin(y)}{2}$
The first case holds when $y\in \{0,\pi\}$ and by simple manipulation and checking the derivatives, the only solution for the second case is $y = \pi$ (of course for $y \in [0,\pi]$).
Thus, as there is no zero for $0\leq y \leq \pi$, either:
$\forall y;  \cos(y) \geq \cos(y + \frac{\sin(y)}{2})$
or
$\forall y;  \cos(y) \leq \cos(y + \frac{\sin(y)}{2})$
For $y = \pi/2, \cos(\pi/2) > \cos(\pi/2 + \frac{1}{2\sqrt{2}}) \Rightarrow \forall y; \cos(y) \geq \cos(y + \frac{\sin(y)}{2})$
Hence, $\forall y\in \mathbb{R}$;
$\cos(2x)(\frac{\cos(2x)+3}{2}) \geq \cos(2x + \frac{\sin(2x)}{2})$
$\blacksquare$
