Minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$ 
If $\alpha,\beta,\gamma$ be a variable and $k$ be a constant such that $a\tan\alpha+b\tan\beta+c\tan\gamma=k$.Then find minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$ is

Try: Using Cauchy Schwarz Inequality:
$(a^2+b^2+c^2)(\tan^2\alpha+\tan^2\beta+\tan^2\gamma)\geq (a\tan\alpha+b\tan\beta+c\tan\gamma)^2$
So we have $$\tan^2\alpha+\tan^2\beta+\tan^2\gamma)\geq \frac{k^2}{a^2+b^2+c^2}$$
Could some help me to solve it without Cauchy Schwarz Inequity . 
Please explain, thanks

Let
\begin{align}
f(x,y,z)&=\tan^2 x + \tan^2 y+\tan^2 z\\
g(x,y,z)&=a\tan x +b\tan y+c\tan z=k
\end{align}
We want to find the minimum of $f$ constrained to $g$, and we can apply Lagrange Multiplier. Consider the equation $\nabla f=\lambda \nabla g$. Since
\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)&=(a\sec^2 x, b\sec^2 y,c\sec^2 z),
\end{align}
for a root of the equation $(x^*,y^*,z^*)$, this equation holds:
$$\begin{cases}\tag 1
2\tan x^*=\lambda a\\
2\tan y^*=\lambda b\\
2\tan z^*=\lambda c
\end{cases}$$
Substitite (1) into $g(x,y,z)=k$, then $\lambda=\dfrac{2k}{a^2+b^2+c^2}$ and
$$
f(x^*,y^*,z^*)=\frac{\lambda^2 (a^2+b^2+c^2)}{4}=\frac{k^2}{a^2+b^2+c^2}.
$$
This is what we want.
