Minimise $\tan^2 A + \tan^2 B + \tan^2 C$ given a relation

Question

Its said that $A,B,C,a$ are real and $\sqrt{a^2-4}\tan A + a \tan B + \sqrt{a^2+4} \tan C = 6a$

then minimise $\tan^2 A + \tan^2 B + \tan^2 C$.

I dont know how to proceed, looked like something to do with pythagoras theorem. Nothing more is given, so any hint is welcome.

Answer 1

For $a^2\geq4$ by C-S we obtain: $$\left(\sqrt{a^2-4}\tan A + a \tan B + \sqrt{a^2+4} \tan C\right)^2\leq$$ $$\leq(a^2-4+a^2+a^2+4)\left(\tan^2 A + \tan^2 B + \tan^2 C\right).$$ Thus, $$\tan^2 A + \tan^2 B + \tan^2 C\geq\frac{36a^2}{3a^2}=12.$$ The equality occurs for $$\left(\sqrt{a^2-4},a,\sqrt{a^2+4}\right)||\left(\tan A,\tan B,\tan C\right),$$ which says that we got a minimal value.

Answer 2

We can use Lagrange multiplier. Let \begin{align} f(x,y,z)&=\tan^2 x + \tan ^2 y + \tan^2 z,\\ g(x,y,z)&=\sqrt{a^2-4}\tan x+a\tan y + \sqrt{a^2+4}\tan z-6a=0. \end{align} Then \begin{align} \nabla f(x,y,z)&=(2\tan x\sec^2 x, 2\tan y\sec^2 y, 2\tan z\sec^2 z),\\ \nabla g(x,y,z)&=(\sqrt{a^2-4}\sec^2 x, a\sec^2 y,\sqrt{a^2+4}\sec^2 z). \end{align} Consider the equation $\nabla f = \lambda \nabla g$, then for a root of the equation $(x^*,y^*,z^*)$, \begin{cases}\tag 1 2\tan x^*=\lambda\sqrt{a^2-4}\\ 2\tan y^*=\lambda a\\ 2\tan z^*=\lambda\sqrt{a^2+4} \end{cases} Substitute (1) into $g(x,y,z)=0$, then we get $\lambda=4/a$ and \begin{align} \tan^2 x^*+\tan^2 y^*+\tan^2 z^* &=\left(\frac{2\sqrt{a^2-4}}{a}\right)^2+\left(\frac{2a}{a}\right)^2+\left(\frac{2\sqrt{a^2+4}}{a}\right)^2\\ &=12. \end{align} Thus 12 is what we want to find.