If $$f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}},x > 0$$, then the least value of $f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$ is :

Given that $f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}$ <br/><br/>Let us consider a similar function of $(x)$, <br/><br/>$\therefore f(x)=\frac{A x+B}{C x-A}$ <br/><br/>$\text { Now, } $ <br/><br/>$$ \begin{aligned} &f(f(x)) =\frac{A\left(\frac{A x+B}{C x-A}\right)+B}{C\left(\frac{A x+B}{C x-A}\right)-A} \\\\ & =\frac{A^2 x+A B+B C x-A B}{A C x+B C-A C x+A^2} \\\\ & =\frac{x\left(A^2+B C\right)}{\left(B C+A^2\right)}=x \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore \quad f(f(x))=x \\\\ & \text { Similarly, } f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x} \\\\ & \text { Apply, AM } \geq \text { GM } \\\\ & \left(x+\frac{4}{x}\right) \geq 2 \sqrt{x \cdot \frac{4}{x}}=4(x>0) \\\\ & \Rightarrow f(f(x))+f\left(f\left(\frac{4}{x}\right)\right) \geq 4 \end{aligned} $$ <br/><br/>Hence, the least value is 4 .