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.
 A: 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 $$d^2=\frac{1600}{98}.$$
A: Define $f:\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)^3\to \Bbb R$ putting $$f(x,y,z) =  \tan^2 x+\dfrac{\tan^2 y}{4}+\dfrac{\tan^2 z}{9} $$
You want to optimize $f$ subject to $$g(x,y,z) = \tan x+2\tan y+3\tan z=40$$
Call $u = \tan x$, $v = \tan y$ and $w = \tan z$. Since $x,y,z \in \left(\frac{\pi}{2},\frac{3\pi}{2}\right)$, we have that $u,v,w \in \Bbb R$.
It seems easier to me work with: $$\tilde{f}(u,v,w) = u^2 + \frac{v^2}{4} + \frac{w^2}{9}$$ and the restriction: $$\tilde{g}(u,v,w) = u + 2v + 3w = 40$$
Now use Lagrange Multipliers and solve: $$\begin{cases} \nabla \tilde{f}(u,v,w) = \lambda \nabla \tilde{g}(u,v,w) \\ \tilde{g}(u,v,w) = 40 \end{cases}.$$
Your method seems fine to me, though.
A: Your method is equivalent to the following
$$
a+2b+3c=40
$$
$$
\text{minimize }a^2+\frac14b^2+\frac19c^2
$$
Cauchy-Schwarz says that
$$
1\cdot a+4\cdot\frac12b+9\cdot\frac13c\le\sqrt{1+4^2+9^2}\sqrt{a^2+\frac14b^2+\frac19c^2}
$$
where equality occurs only when $\left.\left(a,\frac12b,\frac13c\right)\middle\|\left(1,4,9\right)\right.$.
That is,
$$
1600\le98\left(a^2+\frac14b^2+\frac19c^2\right)
$$
and that for some $(a,b,c)$, like $\left(\frac{40}{98},\frac{320}{98},\frac{1080}{98}\right)$, equality holds.
A: Let $$a:=\tan x,b:=\frac12 \tan y,c:=\frac13\tan z$$
Now $a+4b+9c=40$ and we need to minimize $(a^2+b^2+c^2)$
Now using Cauchy Schwarz: $$(a^2+b^2+c^2)(1+16+81)\ge(a+4b+9c)^2$$
