<s>[INST] Showing that $ |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$ For every nonnegative integer $n$ and every real number $ x$ prove the inequality:
$$\sum_{k=0}^n|\cos(2^kx)|= |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$$
 [/INST] This is terse and skips some important steps, but I speculate that it can be turned nicely into a clean proof!
We show:
$$|\cos 2^n x | + |\cos 2^{n+1} x| \geq \frac{1}{\sqrt{2}}$$
We have
$$
\sqrt{\cos^2 2^n x } + \sqrt{\cos^2 2^{n+1} x}.
$$ And since $\cos^2 x = \frac{1}{2}(\cos(2x)+1)$ we get
$$
\frac{1}{\sqrt{2}}\sqrt{\cos (2^{n+1} x) + 1} + \sqrt{\cos^2 (2^{n+1} x)}.
$$ Substituting we get
$$
\frac{1}{\sqrt 2}\sqrt{z+1}+ |z|
$$ which has the minimum $\frac{1}{\sqrt 2}$ at $z=0$.
EDIT: Outline of complete proof
If $n$ is even, we are done since we have $n/2$ pairs.
For $n$ is odd there might be a simple proof, but I couldn't come up with anything better than this.
We show
$$|\cos 2^{n-1} x | + |\cos 2^n x | + |\cos 2^{n+1} x| \geq \frac{3}{2\sqrt{2}}$$
With tricks as above but also using $\cos(2x) = 2\cos^2(x) - 1$ on the last term we get
$$\frac{1}{2}\sqrt{\cos 2^{n} x +1} + |\cos 2^n x | + |2\cos^2 2^{n} x-1| \geq \frac{3}{2\sqrt{2}}$$
and thus
$$
\frac{1} {\sqrt 2}\sqrt{z+1}+ |z| + |2z^2-1|
$$
which has minimum 
$$
\frac{\sqrt{2}+\sqrt{2-\sqrt{2}}}{2} > \frac{3}{2\sqrt{2}}
$$
(Since wolframalpha told me so (difference 0.02913, hard one)).
Now we can use this inequality to get an even number of pairs.
Anyone have a slicker proof for the odd case? This one sucks.
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