The least value of the$f\left(x\right)$given by $f\left(x\right)={\tan }^{-1}x-\frac{1}{2}{\log }_{e}x$ in the interval $[1/\sqrt{3},\sqrt{3}],$ is

We have,

$\begin{array}{l}f\left(x\right)={\tan }^{-1}x-\frac{1}{2}{\log }_{e}x\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{1}{1+x^2}-\frac{1}{2x}\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{2x-1-x^2}{2x\left(1+x^2\right)}\\ \therefore f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow 2x-1-x^2=0\Rightarrow x^2-2x+1=0\Rightarrow x=1.\end{array}$

Now,

$f\left(1\right)={\tan }^{-1}1=\frac{\pi }{4},f\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}-\frac{1}{2}{\log }_e\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}+\frac{1}{4}{\log }_{e}3$

and, $f\left(\sqrt{2}\right)=\frac{\pi }{3}-\frac{1}{4}{\log }_{e}3$

Hence, the least value of $f\left(x\right)$ is $\frac{\pi }{3}-\frac{1}{4}{\log }_{e}3$