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
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A bit of Geometry.
$x=\tan \alpha$; $ y= \tan \beta$; $z=\tan \gamma$.
Plane: $ax+by +cz=k.$
$d^2 := x^2+y^2+z^2$ is the squared distance from the origin to a point $(x,y,z)$.
Need to find the (perpendicular= minimal) distance of the plane from the origin.
$\vec n = \dfrac{1}{a^2+b^2+c^2}(a,b,c)$ is the unit normal of the plane.
$\vec r(t) = t \vec n.$
Plane:
$\vec n\cdot(\vec r(t) -\vec r_0)=0.$
Choose: $\vec r_0 = (0,0,k/c).$
Find $t:$
$\vec n \cdot (t\vec n -\vec r_0)=0$, or
$t - \vec n \cdot \vec r_0=0.$
Hence:
$t= \dfrac{k}{\sqrt{a^2+b^2+c^2}}.$
Since $\vec n$ is a unit vector $t$ is the distance from the origin to the plane .
We are looking for the squared distance:
$t^2 = \dfrac{k^2}{a^2+b^2+c^2}.$
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.