Find the minimum of $S = ax+by+cz$ for $x,y,z > 0$ given that $x+y+z = xyz$ Given $x,y,z \in \Bbb R^+$ satisfying
$$
x+y+z = xyz \tag{1},
$$

find the minimum value of the sum
$$
S=ax+by+cz,
$$
with $a,b,c \in \Bbb R^+$ are known parameters. For example,
$$
\color{red}{(a,b,c) = \left(\sqrt 7, \frac{\sqrt{7}}{2},\frac{2\sqrt{7}}{7}\right)}
$$
The problem looks very simple. It suffices to find a plane $ax+by+cz = S$ which is tangent to the surface $x+y+z = xyz$.
I know there exists a very clever (but unluckily not intuitive):

*

*From $(1)$, there exists a triangle $ABC$ satisfying:
$x = \tan A,y =\tan B$ and $z = \tan C$.


*The sum $S$ becomes  $S = a \tan A+b \tan B + c\tan C $


*Applying the inequality: $\tan x \ge \tan y +(\tan y)'(x-y)$ for $(x,y) =  (A,p), (B,q)$ and $(C,r)$. These values $p,q,r$ are determined such that $a(\tan p)' = b(\tan q)' = c(\tan r)'$ and $p+q+r = \pi$


*The minimum of $S$ is equal to $a \tan p +b\tan q +c\tan r+...$
For $(a,b,c) = \left(\sqrt 7, \frac{\sqrt{7}}{2},\frac{2\sqrt{7}}{7}\right)$, $S$ reaches its minimum of $\color{red}{\frac{15}{2}}$ when $$
\color{red}{(x,y,z) =\left(\frac{3}{\sqrt{7}}, \frac{5}{\sqrt{7}},\sqrt{7}\right)}
$$
I believe there exists a more intuitive solution without using trigonometry.

Here is my attempt (not beautiful yet)
Let denote
$$P = ax+by+cz + \lambda(x+y+z - xyz)$$
From the first derivatives of $P$, we deduce
$$
\begin{cases}
\frac{a}{\lambda}+1 = yz \\
\frac{b}{\lambda}+1 = zx \\
\frac{c}{\lambda}+1 = xy\\  \tag{2}
\end{cases}
$$
From $(1),(2)$, we have a cubic equation of $\lambda$
$$\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy} =1 \implies 
\lambda^3   +\frac{a+b+c}{2}\lambda^2 -\frac{abc}{2} = 0 \tag{3}$$
Remark: At this step, I realize that the solution depends on the parameters $a,b,c$ (and then the equation $(3)$ can have 1 or $3$ roots). For $
\color{red}{(a,b,c) = \left(\sqrt 7, \frac{\sqrt{7}}{2},\frac{2\sqrt{7}}{7}\right)}
$, we have $\lambda = -\sqrt{7}/4$, $\frac{1}{4}(-\sqrt{7}-\sqrt{14})$ or $\frac{1}{4}(-\sqrt{7}+\sqrt{14})$.
The second derivarives of $P$ is a Hessian matrix:
$$\begin{pmatrix}
0 & \lambda z&\lambda y\\
\lambda z &0 & \lambda x\\
\lambda y & \lambda x &0\\
\end{pmatrix} \tag{4}$$
The equation of 3 eigenvalues of $(4)$
$$p^3  - (x^2+y^2+z^2)p-2xyz =0 \tag{5}$$
Remark: It's very strange because the equation $(5)$ can't have $3$ positive roots nor $3$ negative roots, this means that the Hessian matrix can't be definite positive (negative). And all critical points of $(3)$ aren't extremum points of $S = ax+by+cz$.

Where is my error?
 A: Here is a solution without using trigonometry.
Note: when I wrote the conclusion of this method, I found a second method which is simpler. I don't have time to write both two methods here.
Let denote
$$P = ax+by+cz + \lambda(x+y+z - xyz)$$
From the first derivatives of $P$, we deduce
$$
\begin{cases}
\frac{a}{\lambda}+1 = yz \\
\frac{b}{\lambda}+1 = zx \\
\frac{c}{\lambda}+1 = xy\\  \tag{2}
\end{cases}
$$
From $(1),(2)$, we have a cubic equation of $\lambda$
$$\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy} =1 \implies 
\lambda^3   +\frac{a+b+c}{2}\lambda^2 -\frac{abc}{2} = 0 \tag{3}$$
The equation $(3)$ has one positive root. We will prove that this root is unique in $\Bbb R$. Indeed, we know that $(3)$ can't have $3$ positive roots, then we suppose that there exists a negative root $\lambda'$, then
$$yz = 1-\frac{a}{\lambda'}<1 \implies x+y+z = xyz <x \implies y+z<0  \text{ : contradiction}$$
So, there is only one critical point $(x,y,z,\lambda)$ satisfying $(2)$ and $(3)$. ( $(x,y,z,\lambda)$ can be solved analytically).
The second derivarives of $P$ is a bordered Hessian matrix:
$$\begin{pmatrix}
0 & -\frac{a}{\lambda} &-\frac{b}{\lambda} &-\frac{c}{\lambda} \\
-\frac{a}{\lambda}& 0 & -\lambda z & -\lambda y \\
-\frac{b}{\lambda} & -\lambda z &0 & -\lambda x\\
-\frac{c}{\lambda} & -\lambda y & -\lambda x &0 \\
\end{pmatrix} \tag{4}$$
Remark: In my attempt above, I made a mistake by not including the $4$-th dimension with the variable $\lambda$. And the bordered Hessian matrix is not necessary to be definite positive in the case of minimum with constraints.
The point $(x,y,z,\lambda)$ can be proved as a minimum point by checking the sufficient condition for local extrema or here . The idea is to calculate the determinant of $2$ minors of the Hessian matrix. The $2$ determinants must be negative (same sign as $(-1)^1 = -1$).

*

*The first determinant is equal to $-2\frac{abz}{l}<0$

*The second determinant (the determinant of Hessian matrix) is equal to $(a^2x^2+b^2y^2+c^2z^2)-2(abxy+bcyz+cazx)$. I haven't simplified this formula yet but it must be negative as well.

Conclusion: $P$ reaches its minimum equal to $$P_{\text{min}} = 2\frac{x_0y_0z_0}{\lambda}$$
with $(x_0,y_0,z_0,\lambda)$ is the solution of $(2)$ and $(3)$( $(x_0,y_0,z_0,\lambda)$ can be calculated analytically).
Remark: When I was writing the conclusion, I just found a second method which seems simpler than this method. Thanks to the second method, I could have easily the minimum value $P_{\text{min}} = 2\frac{x_0y_0z_0}{\lambda}$ (with the first method above, it's not easy to get $P_{\text{min}}$). I'll write the second method later.
