Minimizing Tangents!

With The Given Equation and The Required Function,

Lets Assume a Vectors $\alpha \space \text{and} \space \beta$ such that,

$\vec{\alpha} = \tan{A}\hat{i} + \tan{B}\hat{j} + \tan{C}\hat{k}$

and

$\vec{\beta} = \sqrt{a^2-4}\hat{i} + a\hat{j} + \sqrt{a^2+4}\hat{k}$

Now, We Need to $\textbf{Minimize} |\vec{\alpha}|^{2}$

And Given That

$\vec{\alpha} \cdot \vec{\beta}$ = $\sqrt{a^2-4}\tan{A} + a\tan{B} + \sqrt{a^2+4}\tan{C}$

And

$\vec{\alpha} \cdot \vec{\beta}$ = $6a$

Now,

$\vec{\alpha} \cdot \vec{\beta}$ = $|\vec{\beta}||\vec{\alpha}|\cos{\theta}$

So,

$|\vec{\beta}||\vec{\alpha}|\cos{\theta}$ = $6a$

And $|\vec{\beta}| = \sqrt{3}a$

So,

$|\vec{\alpha}|$ = $\dfrac{6}{\sqrt{3}\cos{\theta}}$

And As $\cos{\theta}$ Becomes Maximum $|\vec{\alpha}|$ becomes minimum

Therefore,

$|\vec{\alpha}|_{Min}$ = $2\sqrt{3}$

And

$|\vec{\alpha}|^{2}_{Min}$ = $\boxed{12}$

Using Cauchy Schwartz Inequality gives $(6a)^2=<((tanA)^2+(tanB)^2+(tanC)^2)(a^2-4+a^2+a^2+4)$ So the required minima is $12$

$\begin{aligned} \sqrt{a^2-4}\tan A + a \tan B + \sqrt{a^2+4} \tan C & = 6a & \small \color{#3D99F6} \text{Dividing both sides by }a \\ \sqrt{1- \frac 4{a^2}} \tan A + \tan B + \sqrt{1+\frac 4{a^2}} \tan C & = 6 \end{aligned}$

Using Cauchy-Schwarz inequality , we have:

$\begin{aligned} \left(\sqrt{1- \frac 4{a^2}} \tan A + \tan B + \sqrt{1+\frac 4{a^2}} \tan C \right)^2 & \le \left(1- \frac 4{a^2} + 1 + 1+\frac 4{a^2} \right) \left( \tan^2 A + \tan^2 B + \tan^2 C \right) \\ 36 & \le 3 \left( \tan^2 A + \tan^2 B + \tan^2 C \right) \\ \implies \tan^2 A + \tan^2 B + \tan^2 C & \ge \boxed{12} \end{aligned}$

Equality occurs when $\dfrac {\tan A}{\sqrt{1- \frac 4{a^2}}} = \tan B = \dfrac {\tan C}{\sqrt{1+ \frac 4{a^2}}}$