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I want to prove the following inequality: $\ln(\cos x)\ge \dfrac{-x^2}{\cos^2(x)}$

Please I'm stuck with this problem, maybe considering the equivalent inequality $\ln(\sec x)\ge (x \sec x)^2$ would help, but I'm not sure. I want to prove it by the use of the Mean Value theorem.

Trafalgar
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2 Answers2

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Consider $f(x) = x^2 + \cos^2(x) \ln(\cos(x))$. Show that $f(x)$ is increasing for $x \in [0,\pi/2)$ and conclude what you want.

  • The derivative looks untreatable. Is this what you have in mind? – Git Gud Apr 24 '13 at 19:27
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    @GitGud It is actually pretty easy to show that the derivative is positive. Use the fact that $-\sin(x)\cos(x) \log(\cos(x))$ is always positive and $2x - \sin(x) \cos(x) = 2x - \dfrac{\sin(2x)}2 > 0$. –  Apr 24 '13 at 19:31
  • @Inceptio What? Can't you differentiate $\log(\cos(x))$ and $\cos^2(x)$? You do not need wolfram for this. You often seem to leave comments without thinking. –  Apr 24 '13 at 19:34
  • But my teacher wants a proof by the use of the mean value theorem )= – Trafalgar Apr 24 '13 at 19:37
  • @user17762: Sorry, deleted my comment after seeing your edited comment. I agree with you(first part only). – Inceptio Apr 24 '13 at 19:37
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$$ \ln (1+x) = x - \frac{x^2}2 + \frac{x^3}3 - \cdots. $$ $$ \ln\cos x = \ln\left(1 -\frac{x^2}{2} + \frac{x^4}{24}-\cdots\right) \ge\ln\left(1 -\frac{x^2}{2}\right) $$ $$ =\frac{-x^2}{2} -\frac{x^4}{8} + \cdots \ge -\frac{x^2}{1.5} \ge-\frac{x^2}{\cos^2 x}\text{ for $x$ close enough to $0$.} $$

Michael Hardy
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