Let $g\left(x\right)=f\left(x\right)+f\left(1-x\right)$ and $f^{"}\left(x\right)>0,x\in \left(0,1\right)$. If $g$ is decreasing in the interval $\left(0,\alpha \right)$ and increasing in the interval $\left(\alpha ,1\right)$, then ${\tan }^{-1}2\alpha +{\tan }^{-1}\left(\frac{1}{\alpha }\right)+{\tan }^{-1}\left(\frac{\alpha +1}{\alpha }\right)$ is equal to

Given,

$g\left(x\right)=f\left(x\right)+f\left(1-x\right)$ and $f^{"}\left(x\right)>0,x\in \left(0,1\right)$. If $g$ is decreasing in the interval $\left(0,\alpha \right)$ and increasing in the interval $\left(\alpha ,1\right)$,

Now solving,

$g\left(x\right)=f\left(x\right)+f\left(1-x\right)$

Now differentiating both side we get,

$g^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)-f^{\text{'}}\left(1-x\right)$

Differentiating again we get,

$g^{"}\left(x\right)=f^{"}\left(x\right)+f^{"}\left(1-x\right)>0$

So, $g^{\text{'}}\left(x\right)$ is increasing as given $f^{"}\left(x\right)>0$

$\Rightarrow g^{\text{'}}\left(0\right)<g^{\text{'}}\left(1\right)$

$\Rightarrow f^{\text{'}}\left(0\right)-f^{\text{'}}\left(1\right)<f^{\text{'}}\left(1\right)-f^{\text{'}}\left(0\right)$

$\Rightarrow f^{\text{'}}\left(0\right)<f^{\text{'}}\left(1\right)$

Now finding $g^{\text{'}}\left(x\right)=0$ we get,

$\Rightarrow f^{\text{'}}\left(x\right)=f^{\text{'}}\left(1-x\right)$

$\Rightarrow x=1-x$

$\Rightarrow x=\frac{1}{2}$

Now $g^{\text{'}}\left(x\right)$ is positive for $x\in \left(0,\frac{1}{2}\right)$

And $g^{\text{'}}\left(x\right)$ is negative for  $x\in \left(\frac{1}{2},1\right)$

$\therefore \alpha =\frac{1}{2}$

So, 

${\tan }^{-1}2\alpha +{\tan }^{-1}\left(\frac{1}{\alpha }\right)+{\tan }^{-1}\left(\frac{\alpha +1}{\alpha }\right)$

$={\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3$

$=\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\left(\frac{2+3}{1-6}\right)$

$=\frac{\mathrm{\pi }}{4}+{\tan }^{-1}\left(-1\right)$

$=\frac{\mathrm{\pi }}{4}+\frac{3\mathrm{\pi }}{4}=\mathrm{\pi }$