Symmetry argument in finding the extrema of $f(x,y,z)=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ that I don't know why works 
Consider the function $$f(x,y,z)=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ and consider the problem to find its extrema in its domain.


I noticed that $f$ is unbounded above and below, so we will search for local extrema: another thing I noticed is that exchanging $y$ with $x$ and exchanging $z$ with $x$ doesn't change the function, that is $f(x,y,z)=f(y,x,z)=f(z,y,x)$, so there is some kind of symmetry in the function $f$.
So, since interchanging the first and the second coordinate and the first and the third coordinate doesn't change $f$, I thought that maybe I could study the function $g(x):=f(x,x,x)=3x+\frac{3}{x}$ to get some informations about $f$. Since
$$g'(x)=3-\frac{3}{x^2} =\frac{3x^2-3}{x^2} = 0 \iff x = 1 \vee x = -1$$
So $g(1)=f(1,1,1)=6$ and $g(-1)=f(-1,-1,-1)=-6$ are extrema for $f$, and according to Wolfram|Alpha this thing is actually true because $f$ reach its local minimum at $(1,1,1)$ and its local maximum at $(-1,-1,-1)$; the problem is that I don't know why this works, and actually I'm not even sure if this has some sense or it is just a casualty.
Can someone explain me what is going on here and, if this has sense, how to make this reasoning rigorous explaining why this works? Thanks.
 A: To answer your question, it is due to the 1) linearity and 2) commutative property of addition operator.
In a slightly more abstract case, consider a function $K:\mathbb{R}^2\mapsto \mathbb{R}$ is both 1) commutative (i.e. $K(x,y)=K(y,x)$) and 2) linear (i.e. $\alpha K(x,y) = K(\alpha x, \alpha y)$ and $K(x+y)=K(x)+K(y)$). Then, from the commutative property, apparently we have:
$$
J = K(x,y)\\
J = K(y,x)
$$
By adding these two equations together and dividing both sides by a factor of $2$, we will have:
$$
J = \frac{1}{2}\left( K(x,y) + K(y,x) \right)=K\left(\frac{1}{2}(x+y), \frac{1}{2}(x+y) \right) =K(p,p)
$$
where $p=(x+y)/2$. Then, the optimisation problem of function $K$ can be reduced into a lower-dimension problem of optimising $R(p)$ with $R(p)=K(p,p)$. Following this line, you can easily generalise the process to $n$-dimensional problem.
Now, back to your case, if you let $u=x+x^{-1}$, $v=y+y^{-1}$, and $w=z+z^{-1}$, the original problem of finding extrema can be transformed into a simpler but equivalent problem $f(x,y,z) = g(u,v,w)=u+v+w$ with $u,v,w\geq C$ and constant $C$, which apparently satisfies both the linearity and commutativity. That is how it works.
A: Might be helpful-
In a symmetric function, will equality always imply a maxima or minima?
For this case, the points of extrema lied on the line x=y=z and so that resulted in above.
But let's consider a simpler case of f(x, y, z) = x + y + z,
if we assume the above theory to be true for some time, then
x + x + x = 0
=> 3x = 0
=> x = 0 is a point of extrema for the given function, but it's actually not as the function is unbounded.
