There's the problem:
given that $0<\alpha<\beta<\frac{\pi}{2}$ , what is bigger
$$\frac{\ln(\cos\alpha)-\ln(\cos\beta)}{\beta-\alpha}$$
or
$$\tan(\frac{\alpha+\beta}{2})$$
I tried to simplify a little bit the first term. Now its $$\frac{\ln(\frac{\cos\alpha}{\cos\beta})}{\beta-\alpha}$$.
Than I tried raise $e$ to the power of both sides and see If I can prove what's bigger, but it was not working. Can someone help please?
Thanks!
One possible way would be to perform Taylor expansions built at $\beta=\alpha$ assuming that $\beta-\alpha$ is small.
This would give, after trigonometric simplifications, $$A=\frac{\log (\cos (\alpha ))-\log (\cos (\beta ))}{\beta -\alpha }=\tan (\alpha )+\frac{1}{2} (\beta -\alpha ) \sec ^2(\alpha )+\frac{1}{3} (\beta -\alpha )^2 \tan (\alpha ) \sec ^2(\alpha )+O\left((\beta -\alpha )^3\right)$$ $$B=\tan \left(\frac{\alpha +\beta }{2}\right)=\tan (\alpha )+\frac{1}{2} (\beta -\alpha ) \sec ^2(\alpha )+\frac{1}{3} (\beta -\alpha )^2 \tan (\alpha ) \sec ^2(\alpha )+O\left((\beta -\alpha )^3\right)$$
Now the difference between the first and the second is $$\Delta=A-B=\frac{1}{12} (\beta -\alpha )^2 \tan (\alpha ) \sec ^2(\alpha )+O\left((\beta -\alpha )^3\right)$$ and since $\beta > \alpha$, then $\cdots$