Find the minimum value of $f$ 
Find the minimum value of $$f(x)=\frac{\tan \left(x+\frac{\pi}{6}\right)}{\tan x}, \qquad x\in \left(0,\frac{\pi}{3}\right).$$

My approach is as follows. I tried to solve it by segregating it  
$$f(x)=\frac{1}{\sqrt{3}\tan x}+\left({\sqrt{3}}+\frac{1}{\sqrt{3}}\right)\frac{1}{\sqrt{3}-\tan x},$$ but $f'(x)$ is getting more and more complicated.

Let me prove a slightly more general result: 

If $a\in (0,\pi/2)$, the function $$f(x) = \frac{\tan(x+a)}{\tan x}$$ has a unique minima at $x=b/2$ for $x\in(0,b)$ where $b=\pi/2-a$.

Proof: First of all note that the function $f$ is symmetric with respect to $b$, i.e., $f(x)=f(b-x)$, since $$f(b-x)=\frac{\tan(\pi/2-x)}{\tan(\pi/2-a-x)}=\frac{\tan(a+x)}{\tan x}.$$ Therefore, it suffices to study the function only for $x\in (0,b/2)$. Now note that the function $f$ can be expressed as $f(x) = g(\tan x)$, where $$g(u) = \frac{\alpha+u}{u(1-\alpha u)},$$ where $\alpha=\tan a$. Now, $$g'(u) = \frac{u(1-\alpha u)-(\alpha+u)(1-2\alpha u)}{u^2(1-\alpha u)^2}=\frac{\alpha((u+\alpha)^2-(1+\alpha^2))}{u^2(1-\alpha u)^2}.$$ Since we consider only $x\in (0,b/2)$, we have, $u=\tan x\le \tan (b/2)=\frac{\sqrt{1+\cot^2 b} - 1}{\cot b}=\sqrt{1+\alpha^2}-\alpha$, which implies that $g'(u)\le 0$, i.e., $g$ is decreasing as long as $u\in (0,\tan(b/2))$. Therefore, when $x\in (0,b/2)$, the function $f(x)=g(\tan x)$ is decreasing in $(0,b/2)$. Hence, the unique minima is obtained at $x=b/2\ \blacksquare$.
