<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] Actually, the statement is true even when one restrict the sum to start at $k=1$.
i.e. For all $x \in \mathbb{R}$,
$$\sum_{k=1}^n |\cos(2^k x)| \ge \frac{n}{2\sqrt{2}} \Leftrightarrow \varphi_n(2x) \ge 0$$
where $\varphi_n(x) = \sum_{k=0}^{n-1}|\cos(2^kx)| - \frac{n}{2\sqrt{2}}$.
Notice:
$$\begin{align}
\varphi_4(x) &= \varphi_2(x) + \varphi_2(4x)\\
\varphi_5(x) &= \varphi_2(x) + \varphi_3(4x)\\
&\,\,\vdots\\
\varphi_n(x) &= \varphi_2(x) + \varphi_{n-2}(4x)
\end{align}$$
One only need to show $\varphi_2(x), \varphi_3(x) \ge 0$ for all $x$.
As other answer suggested, $|\cos(2^k x)|$ are piecewise concave functions. We have:
$$\frac{d^2}{dx^2}|\cos(2^k x)| = -4^k |\cos(2^k x)| < 0 \,\,\,\text{ for } 
x \ne \pm 2^{-k}(l+\frac12)\pi \text{ where } l \in \mathbb{N}
$$
This in turn implies
$$\frac{d^2}{dx^2}\varphi_n(x) < 0 \,\,\,\text{ for } x \ne \pm2^{-n}(l+1)\pi \text{ where } l \in \mathbb{N}$$ 
The absolute minimum of $\varphi(x)$ must be located at a $x_{min}$ of the form $\pm 2^{-n}(l+1)\pi$.
Since $\varphi(x)$ is even and has period $\pi$, we only need to look for $x_{min}$ over the interval $[0, \frac{\pi}{2}]$.
For all x, we have:
$$\begin{align}
\varphi_2(x) &\ge \min(\varphi_2(\frac{\pi}{4}),\varphi_2(\frac{\pi}{2})) = \varphi_2(\frac{\pi}{4}) = 0\\
\varphi_3(x) &\ge \min(\varphi_3(\frac{\pi}{8}), \varphi_3(\frac{\pi}{4}), \varphi_3(\frac{3\pi}{8}), \varphi_3(\frac{\pi}{2})) = \varphi_3(\frac{3\pi}{8}) = \cos(\frac{3\pi}{8}) - \frac{1}{2\sqrt{2}} > 0
\end{align}$$
</s>