Inequality exercise asked by Dan Sitaru in many math groups

Exercise:

Show that, $2\left( \arctan \pi -\arctan e \right)<\ln \pi -1$

Solution: Let $f\left( x \right)=2\arctan x-\ln x$ be function continuous on $\left[ e,\pi  \right]$ and differentiable on

$\left( e,\pi  \right)$ by MVT there exists $c\in \mathbb{R}\,\,,\,\,e\le c\le \pi $ such that $f'\left( c \right)=\frac{f\left( \pi  \right)-f\left( e \right)}{\pi -e}$

Observe that $f'\left( c \right)=\frac{2}{1+{{c}^{2}}}-\frac{1}{c}=\frac{2c-1-{{c}^{2}}}{c\left( 1+{{c}^{2}} \right)}=\frac{-\left( {{c}^{2}}+1-2c \right)}{c\left( 1+{{c}^{2}} \right)}=\frac{-{{\left( c-1 \right)}^{2}}}{c\left( 1+{{c}^{2}} \right)}<0$

Hence $\frac{f\left( \pi  \right)-f\left( e \right)}{\pi -e}<0\Leftrightarrow f\left( \pi  \right)<f\left( e \right)\Leftrightarrow 2\arctan \pi -\ln \pi <2\arctan e-1$

$\Leftrightarrow 2\left( \arctan \pi -\arctan e \right)<\ln \pi -1$

Or: take $f\left( x \right)=2\arctan x-\ln x\,\,,\,\,x>0$ be a continuous function

So $f'\left( x \right)=\frac{2}{1+{{x}^{2}}}-\frac{1}{x}=\frac{2x-1-{{x}^{2}}}{x\left( 1+{{x}^{2}} \right)}=\frac{2x\left( 1-x \right)-1}{x\left( 1+{{x}^{2}} \right)}$

But $x>0\Leftrightarrow -x<0\Leftrightarrow 1-x<1$ so $f'<0$ hence $f$ is decreasing function

But \[1\le e\le \pi \Leftrightarrow f\left( \pi  \right)\le f\left( e \right)\le f\left( 1 \right)\Leftrightarrow 2\arctan \pi -\ln \pi \le 2\arctan e-1\le \frac{\pi }{4}\]

Hence $2\arctan \pi -\ln \pi \le 2\arctan e-1\Leftrightarrow 2\left( \arctan \pi +\arctan e \right)\le \ln \pi -1$