Minimum of a three variable function (2) This question is related to the question here.The referred question was about finding mininum/maximum of $\frac{x^2+y^2+z^2}{xyz}$.

The answers were clear. But one of the hint given by @Marconius is given below

"Also try to exploit the symmetry of the function, i.e. if you can prove that a minimum can only occur when $x=y=z$, then you can reduce the problem to a single-variable one"

My question is that how such a proof can be derived. Could anyone provide me right directions. For example, can we prove that $\frac{x^2+y^2+z^2}{xyz}$ has a minimum when $x=y=z$ (for positive real values of $x,y,z$)?
The graph of the function $\tfrac{x^2+y^2+z^2}{xyz}$ is the subset of $\Bbb{R}^4$ given by
$$\left\{(x,y,z,w)\in\Bbb{R}^4\mid\ w=\tfrac{x^2+y^2+z^2}{xyz}\right\}.$$
If you can prove that a minimum/maximum can only occur if $x=y=z$, then any minimum/maximum must be in the subset
$$\left\{(x,y,z,w)\in\Bbb{R}^4\mid\ w=\tfrac{x^2+y^2+z^2}{xyz}\text{ and }x=y=z\right\}=\left\{(x,x,x,w)\in\Bbb{R}^4\mid\ w=\tfrac{3}{x}\right\}.$$
In particular $x$ is then a minimum/maximum of the function $\tfrac{3}{x}$. This reduces the problem to a single-variable problem.

In this particular case the function $\tfrac{3}{x}$ has no minimum/maximum, so the original function has no minimum/maximum with $x=y=z$, and the comment is of no use here.



In general, if you can prove that for a minimum/maximum $m=(x_1,\ldots,x_n)$ of a multivariate function $f(x_1,\ldots,x_n)$ we have an explicit relation $x_n=g(x_1,\ldots,x_{n-1})$, then we can substitute $x_n$ to look for a minimum/maximum of
$$f(x_1,\ldots,x_{n-1},g(x_1,\ldots,x_{n-1})),$$
making it a problem in one variable less. In this way every relation we find reduces the number of variables by one. In this particular case we are looking at
$$f(x,y,z)=\frac{x^2+y^2+z^2}{xyz},$$
and the hint given by @Marconius suggests your prove that the relations
$$z=y\qquad\text{ and }\qquad y=x,$$
both hold. This reduces the problem to finding a minimum/maximum of $f(x,x,x)$, which is a single-variable function. However, the hint is misleading because your function has no minima or maxima at all.
The graph of the function $\tfrac{x^2+y^2+z^2}{xyz}$ is the subset of $\Bbb{R}^4$ given by
$$\left\{(x,y,z,w)\in\Bbb{R}^4\mid\ w=\tfrac{x^2+y^2+z^2}{xyz}\right\}.$$
If you can prove that a minimum/maximum can only occur if $x=y=z$, then any minimum/maximum must be in the subset
$$\left\{(x,y,z,w)\in\Bbb{R}^4\mid\ w=\tfrac{x^2+y^2+z^2}{xyz}\text{ and }x=y=z\right\}=\left\{(x,x,x,w)\in\Bbb{R}^4\mid\ w=\tfrac{3}{x}\right\}.$$
In particular $x$ is then a minimum/maximum of the function $\tfrac{3}{x}$. This reduces the problem to a single-variable problem.

In this particular case the function $\tfrac{3}{x}$ has no minimum/maximum, so the original function has no minimum/maximum with $x=y=z$, and the comment is of no use here.



In general, if you can prove that for a minimum/maximum $m=(x_1,\ldots,x_n)$ of a multivariate function $f(x_1,\ldots,x_n)$ we have an explicit relation $x_n=g(x_1,\ldots,x_{n-1})$, then we can substitute $x_n$ to look for a minimum/maximum of
$$f(x_1,\ldots,x_{n-1},g(x_1,\ldots,x_{n-1})),$$
making it a problem in one variable less. In this way every relation we find reduces the number of variables by one. In this particular case we are looking at
$$f(x,y,z)=\frac{x^2+y^2+z^2}{xyz},$$
and the hint given by @Marconius suggests your prove that the relations
$$z=y\qquad\text{ and }\qquad y=x,$$
both hold. This reduces the problem to finding a minimum/maximum of $f(x,x,x)$, which is a single-variable function. However, the hint is misleading because your function has no minima or maxima at all.