MHB Is it Possible to Prove this Trigonometric Inequality?

[lfdahl]

Prove the inequality:

\[\left | \cos x \right |+ \left | \cos 2x \right |+\left | \cos 2^2x \right |+...+ \left | \cos 2^nx \right |\geq \frac{n}{2\sqrt{2}}\]

- for any real x and any natural number, n.

 

[lfdahl]

Hint:

Spoiler

What is the range of the function:$f(x) = |\cos x| + |\cos 2x|$?

 

[lfdahl]

Suggested solution:

Spoiler

Note, that:

\[\left | \cos x \right |+\left | \cos 2x \right |=\left | \cos x \right |+\left | 2\cos ^2x-1 \right | \equiv t + \left | 2t^2-1 \right |\geq \frac{1}{\sqrt{2}}, \: \: \: \: 0\leq t\leq 1.\]

For odd values of $n$, we get $k+1$ pairs ($n = 2k+1$):
\[\left ( \left | \cos x \right |+\left | \cos 2x \right | \right )+\left ( \left | \cos 4x \right |+\left | \cos 8x \right | \right )+ ...+\left ( \left | \cos 2^{n-1}x \right |+\left | \cos 2^{n}x \right | \right )\geq \frac{k+1}{\sqrt{2}}=\frac{n+1}{2\sqrt{2}} > \frac{n}{2\sqrt{2}}\]

For even values of $n$ ($n = 2k+2$):

\[\left | \cos x \right |+\left ( \left | \cos 2x \right | + \left | \cos 4x \right |\right )+ \left ( \left | \cos 8x \right | +\left | \cos 16x \right | \right ) ...+\left ( \left | \cos 2^{n-1}x \right |+\left | \cos 2^{n}x \right | \right ) \geq =\frac{n}{2\sqrt{2}}\]