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How to prove that $$(\sin x)^{\sin x}<(\cos x)^{\cos x}$$ if x is from $]0, \pi/4[$?

Sierra Res
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    Welcome to MSE! Can you [edit] your question to include your thoughts and efforts on this problem? What have you tried, and where are you having difficulty? This will help people write an appropriate answer the addresses your problem. Questions that include this information tend to have a much better response. –  Jan 04 '16 at 05:37
  • Hey,i have no clue how to write mathematical expressions on the "ask question" category. – Sierra Res Jan 04 '16 at 05:41
  • Learn mathJax in this case: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Olimjon Jan 04 '16 at 05:49
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    Simple : $\sin x-\cos x=\sqrt2\sin(x-\pi/4)$ – lab bhattacharjee Jan 04 '16 at 05:50
  • What sort of proof do you want; i.e.. What field of analysis? – Brevan Ellefsen Jan 04 '16 at 05:55
  • One that uses arithmetic and geometric mean. – Sierra Res Jan 04 '16 at 05:56
  • @labbhattacharjee I don't understand your comment. – David Jan 04 '16 at 08:05
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    Transform the inequality to $\cos x \log \cos x - \sin x \log \sin x > 0$, and after some work, to $\log \cot x > (\cot x - 1)\log \sec x$. Then set $t = \cot x > 1$, and the inequality becomes $2 \log t > (t-1)\log (1 + 1/t^2)$. Now use the inequality $\log v \leq (v - 1)$ to prove $(t - 1)\log(1 + 1/t^2) \leq (t - 1)/t^2 \leq 1/4$. Finally prove $(t - 1)\log(1 + 1/t^2) \leq \min(1/4, (t - 1)\log 2) < 2 \log t$ for $t > 1$. – David Jan 04 '16 at 08:24

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