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.
 A: The first derivate of your function is:
$$f'(x)=\cot(x)\sec(x+\pi/6)^2-\tan(x+\pi/6)\csc(x)^2$$
Now, we have to impose $f'(x)=0$ and so:
$$\sin(x)\cos(x)-\sin(x+\pi/6)\cos(x+\pi/6)=0$$
That can be rewritten as:
$$\sin(2x)=\sin(2x+\pi/3)$$
We know that: $$\sin(\alpha)=\sin(\beta) \leftrightarrow \alpha+\beta=\pi+2k\pi \vee \alpha=\beta+2k\pi$$So we have:
$$2x=2x+\pi/3+2k\pi\leftrightarrow\text{IMPOSSIBLE} \vee
4x+\pi/3=\pi+2k\pi \leftrightarrow x=\pi/6+k\pi/2$$
The only solution is $x=\pi/6$. So:
$$f(\pi/6)=\frac{\sqrt3}{\frac{1}{\sqrt3}}=3$$
A: $$g(x)=1+\dfrac1{2\sin x\cos(x+\pi/6)}=1+\dfrac1{\sin(2x+\pi/6)-\sin\pi/6}$$
Now we need to maximize $$\sin(2x+\pi/6)$$ in $$(\pi/6,2\pi/3+\pi/6)$$
A: We know that $\tan(x)\tan(y)=1$ then $x+y=\frac{\pi}{2}$ thus we have $f(x)=\frac{1}{\tan(x)\tan(\frac{\pi}{3}-x)}$ thus we need to maximize denominator (lll be referring to it as k)to get minimum value . Now using the fact that $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ we have $\tan(\frac{\pi}{3})=\frac{\tan(\frac{\pi}{3}-x)+\tan(x)}{1-k}$ thus $k=g(x)=1-\frac{1}{\sqrt{3}}(tan(\frac{\pi}{3}-x)+\tan(x))$ thus maximum value is at $g'(x)=\sec^2(\frac{\pi}{3}-x)-\sec^2(x)=0$ in the given domain we have only one solution at $x=\frac{\pi}{6}$ thus the minimum value of $f(x)=3$
A: 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 $x=b/2\ \blacksquare$.
A: Using the quadratic equation method
In the following let $-\infty <\tan x =z  < \infty \implies z \in R $
$$y=\frac{\tan(x+1/3)}{\tan x}\implies yz^2+\sqrt{3}(1-y)z+1=0$$
The condition that this quadratic has real roots will determine all possible values of $y$
We demand $B^2 \ge 4AC$, then
$$3(1-y)^2 \ge 4y \implies 3y^2-10y+3 \ge 0 \implies (y-3)(3y-1)\ge 0 \implies y\le 1/3 ~or~y\ge 3$$ Finally $$y \ge 3~~for~~ x~~~\in (0.\pi/3)~~~(1)$$
and $$y\le 1/3, ~~for~~ x\in(\pi/3,\pi)~~~(2)$$
FInally. Eq(1) gives the answer here.
