either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP

Question

Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:

$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$

holds

I don't see room to use derivatives, since it seems a little messy to calculate the $\lim_{x\to 0}$ of $\frac{2\ln(\cos{x})}{x^2}$ (which, I think, is necessary in order to make usage of derivatives). Any hints? I've already tried cross multiplying but it doesn't lead anywhere, unless I missed something.

Answer 1

Note that since $e^x$ is strictly increasing

$$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1\iff\ln \cos x< \frac{x^4}{24}- \frac{x^2}{2}\iff \cos x<e^{\frac{x^4}{24}- \frac{x^2}{2}}$$

and since

• $e^x>1+x$
• $\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$

(refer to Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$)

we have

$$e^{\frac{x^4}{24}- \frac{x^2}{2}}>1-\frac{x^2}{2}+\frac{x^4}{24}>\cos x$$

Answer 2

You need the following two inequalities: $$\cos x = 1- \frac{x^2}{2}+ \frac{x^4}{24}-\frac{x^6}{6!} + \cdots < 1- \frac{x^2}{2}+ \frac{x^4}{24}$$ and $$\ln (1+y) = y -\frac{y^2}{2}+ \cdots\le y$$

(you can formally prove them using derivatives, if you want).

Combining them substituting $y= - \frac{x^2}{2}+ \frac{x^4}{24}$ you get

$$\ln \cos x < \ln \left(1- \frac{x^2}{2}+ \frac{x^4}{24} \right) \le - \frac{x^2}{2}+ \frac{x^4}{24}$$

Multiply both sides by $\frac{2}{x^2}$ to get $$\frac{2 \ln ( \cos x)}{x^2} < -1+\frac{x^2}{12}$$

Answer 3

First note that $\sec^2 x > 1$ if $x \in (0,\frac \pi 2)$, so that $$\tan x = \int_0^x \sec^2 t \, dt > \int_0^x 1 \, dt = x$$ for all $x \in (0,\frac \pi 2)$. Similarly, $$ - \ln (\cos x) = \int_0^x \tan t \, dt > \int_0^x t \, dt = \frac{x^2}{2}$$ for all $x \in (0,\frac \pi 2)$. Consequently $$\frac{2 \ln(\cos x)}{x^2} < -1$$ for all $x \in (0,\frac\pi 2)$. This clearly implies the desired inequality.

Answer 4

The Maclaurin series of the tangent function has all coefficients positive (see formula). Integrating we get that the Maclaurin series of the function $-\log(\cos x)$ has all coefficients positive. Up to order $12$ it is $$-\log (\cos x) =\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31 x^{10}}{14175} + \frac{691 x^{12}}{935550} + \mathcal{O}(x^{14})$$ So the inequality should probably read $$-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) >0$$ or $$ \frac{2 \log(\cos(x))}{x^2}<-1 - \frac{x^2}{6}$$

ADDED:

We can prove the inequality $-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) >0$ by taking the derivatives of the function $-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) $ up to order $5$. One can check easily that they all take value $0$ at $0$. Then one calculates: $$\left(-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) \right)^{(5)}=4(5-\cos 2x)\sec^4 x \tan x>0$$ for $0<x<\pi/2$. From here one obtains the inequality . Note that we also get $\tan x>x + \frac{x^3}{3}$.