Minimizing $\tan^2 x+\frac{\tan^2 y}{4}+\frac{\tan^2 z}{9}$ 
Given that $\tan x+2\tan y+3\tan z=40 , \ \ \ x,y,z \in \left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right),$ 
We need to find the minimum value of $ \tan^2 x+\dfrac{\tan^2 y}{4}+\dfrac{\tan^2 z}{9}$

One way would be to consider the vectors 
$$\begin{align}
&\vec{v_1}=\tan x \ \hat{i}+\dfrac{\tan y}{2} \ \hat{j}+\dfrac{\tan z}{3} \ \hat{k} \ \ \ \ \ \ \ \ \text{and}\\ &\vec{v_2}=\hat{i}+4 \ \hat{j}+9 \ \hat{k}
\end{align}$$
And then apply to them the inequality $\left(\vec{v_1} \cdot \vec{v_2}\right)^2 \leq \left|\vec{v_1}\right|^2 \left|\vec{v_2}\right|^2$.
Using this we get $\tan^2 x+\dfrac{\tan^2 y}{4}+\dfrac{\tan^2 z}{9} \geq \dfrac{1600}{98}$
Is this method fine?
What are other methods to solve this problem, which do not involve vectors or complex numbers? Can this result be obtained by using elementary calculus?
Thank you.

The $\tan$ are allowed to span the whole real range, so we can define
$$X=\tan x,Y=\frac12\tan y,Z=\frac13\tan z.$$
The planar constraint becomes
$$X+4Y+9Z=40,$$
and we must find the point closest to the origin (minimize $X^2+Y^2+Z^2$).
This point is obviously on the normal to the plane through the origin, let $(X,Y,Z)=\frac d{\sqrt{98}}(1,4,9)$. Plugging into the plane equation, we have
$$\frac d{\sqrt{98}}(1\cdot1+4\cdot4+9\cdot9)=40,$$
and the requested squared distance is $$d^2=\frac{1600}{98}.$$
