Minimum of $\tan x+\tan y+\tan z$ using Lagrange multipliers for acute angled triangle $\Delta xyz$

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Consider an acute angled triangle $\Delta xyz$. Suppose we want to minimise the function $f(x,y,z)=\tan x+\tan y+\tan z$. So as per the method of Lagrange multipliers here is what I did.

$$\begin{aligned}&\text{minimise}: f(x,y,z)=\tan x +\tan y+\tan z \\ &\text{given: }x+y+z=\pi \end{aligned}$$


$$\begin{aligned}\sec^2 x&=\lambda\\ \sec^2 y&=\lambda \\ \sec^2 z&=\lambda\end{aligned}$$

Now we have the constraint $x+y+z=\pi$ and $0\lt x,y,z\lt \pi/2$. Also do have $4$ unknowns and $4$ equations to solve for $x,y,z,\lambda$. But I am not clear on how to use these equations, managing the $\pm$ signs when we take square roots. Besides if it was not given that $\Delta xyz$ is acute angled, then the minimum would have been in the case of a degenerate triangle when any one of the angles is $0$ and others add up to $\pi$, in which case, $f(x,\pi-x,0)=0$, but I do not understand the jump from $0$ to $3\sqrt{3}$ either for the minimum. Can it not attain values in $(0,3\sqrt{3})$? Why? . Any hints are appreciated. Thanks.

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To handle this problem with the Lagrange Multipliers method, we need first of all, convert the inequalities into equations so with the help of slack variables $(s_x,s_y,s_z,e_x,e_y,e_z)$ the formulation could be

$$ \min_{x,y,z} \tan x + \tan y + \tan z\ \ \ \text{s. t.}\cases{x+y+z=\pi\\ x = s_x^2\\ y = s_y^2\\ z = s_z^2\\ \frac{\pi}{2}-x = e_x^2\\ \frac{\pi}{2}-y = e_y^2\\ \frac{\pi}{2}-x = e_z^2\\ } $$

and the corresponding lagrangian

$$ L = \tan x + \tan y + \tan z + \lambda(x+y+z-\pi)+\sum_{k=\{x,y,z\}} \eta_k(k-s_k^2)+ \sum_{k=\{x,y,z\}} \mu_k\left(\frac{\pi}{2}-k-e_k^2\right) $$

so the set of stationary points for $L$ could contain the solution. Now the minimization problem has the variables $\{x,y,z,\lambda,\eta_k,\mu_k,s_k,e_k\},\ \ k=\{x,y,z\}$. Also, to avoid algebraic problems to obtain the solutions for equations involving trigonometric functions, we can use

$$ \tan x\approx x + \frac{x^3}{3} $$