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Prove:

$$\frac{2\ln(\cos x)}{x^2}<\frac{x^2}{12}-1$$ for $$x \in (0,\frac{\pi}{2})$$

I tried regular derivative methods to prove this. I thought a while about using Taylor series, but without any success. Any hints would be greatly appreciated.

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  • $\begingroup$ Have you tried to find Taylor series for the left hand side of the inequality? $\endgroup$
    – Mhenni Benghorbal
    Mar 20, 2016 at 20:22
  • $\begingroup$ I did. I do noth have much experience with Taylor expansion, so maybe I just do not see what should be obvious. But from comparing it, I only could figure out that this inequality is true when x is close to 0. I am not sure how to proceed to tell something more about it on that basis. $\endgroup$
    – SekstusEmpiryk
    Mar 20, 2016 at 20:29

2 Answers 2

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$f(x) = 2\ln(\cos x) - \dfrac{x^4}{12} + x^2\Rightarrow f'(x) = -2\tan x - \dfrac{x^3}{3}+2x\Rightarrow f''(x) = -2\sec^2 x - x^2 + 2 = -2\tan^2 x - x^2 < 0 \Rightarrow f'(x) < f'(0) = 0 \Rightarrow f(x) < f(0) = 0$. QED.

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By the Leibniz theorem on alternating sequences $$ \cos x < 1-\frac12x^2+\frac1{24}x^4. $$ Since $1+y\le e^y\implies \ln(1+y)\le y$ it follows that $$ 2\ln\cos x\le-x^2+\frac1{12}x^4 $$ which implies the claim of the task.

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