**Motivation**: 


The following problem has occurred in a study of energy dissipation in a chain of coupled, damped oscillators. 


**The problem**: 


Let me define specific rational functions $f$, $g$, and $h$ from $\mathbb R\_+^n$ into $\mathbb R\_+$ ($n \geq 2$) by the following expressions: 
\begin{align\*}
f(x)
& = 1 + \frac{x\_2}{x\_1} + \frac{x\_2 x\_4}{x\_1 x\_3} + \cdots
 = 1 + \frac{x\_2}{x\_1} \left( 1 + \frac{x\_4}{x\_3} \left( 1 + \frac{x\_6}{x\_5} \cdots \right) \right)
\ , 
\tag\*{(1a)}
\\
g(x)
& = 1 + \frac{x\_3}{x\_2} + \frac{x\_3 x\_5}{x\_2 x\_4} + \cdots
 = 1 + \frac{x\_3}{x\_2} \left( 1 + \frac{x\_5}{x\_4} \left( 1 + \frac{x\_7}{x\_6} \cdots \right) \right)
\ , 
\tag\*{(1b)}
\\
\\
h(x)
& = x\_1 f(x)\, g(x)
\\
& = x\_1 + \cdots + x\_n + \frac{x\_1 x\_3}{x\_2} + \frac{x\_2 x\_4}{x\_3} + \cdots + \frac{x\_{n-2} x\_n}{x\_{n-1}}
\\
& \qquad + \frac{x\_1 x\_3 x\_5}{x\_2 x\_4} + \frac{x\_2 x\_4 x\_6}{x\_3 x\_5} + \cdots + \frac{x\_{n-4} x\_{n-2} x\_n}{x\_{n-3} x\_{n-1}}
\\
& \qquad + \frac{x\_1 x\_3 x\_5 x\_7}{x\_2 x\_4 x\_6} + \cdots \quad \text{ etc. }
\ .
\tag\*{(1c)}
\end{align\*}
The dots in (1a-c) indicate a continuation of the respective patterns, with as many terms as possible as long as the indices $i$ on all involved $x\_i$ remain within $\{1,...,n\}$. 


I am interested in a meaningful "minimization" of the function $h$. From its definition above it is obvious that $h$ is both strictly positive and a homogeneous function of degree 1. For any $t\in\mathbb R\_+$ then $h(t\,x) = t\,h(x)$, which shows that $h$ has arbitrarily small values close to the origin of $\mathbb R^n$, and so it cannot have a global minimum in $\mathbb R\_+^n$. An obvious way around this problem is to impose an additional constraint on $x$. Because of the homogeneity of $h$, the natural choice is a constraint of the form 
\begin{align\*}
\prod\_{i=1}^n x\_i = c
\ , 
\tag\*{(2)}
\end{align\*}
with some constant $c>0$. Without loss of generality, we can choose any fixed value, for instance $c=1$; function values for any $c>0$ are (by homogeneity of $h$) related to those for $c=1$ by $h(t\,x) = t\,h(x)$, with $t=c^{1/n}$. 


We know that, on the hypersurface $\mathbb H\_c^n = \{ x \in \mathbb R\_+^n \! : \prod\_{i=1}^n x\_i = c \}$ defined by the constraint (2), $h$ must have a minimum since it is continuous (even $C^\infty$) in $\mathbb R\_+^n$ and (*Proposition 1*:) if $x\_i \to \infty$ for some $i$, or if $x \in \mathbb H\_c^n$ approaches the boundary of $\mathbb R\_+^n$, then $h(x) \to \infty$. Proposition 1 is easily proved as follows. Noting that $h(x) > x\_1 + \cdots + x\_n$, the first part is obvious. For the second part, let us assume that $x\_i < \epsilon$ for some $i$. Then due to the constraint (2) there is at least one $j$ such that $x\_j > (c/\epsilon)^{\frac1{n-1}}$, whence $h(x) > (c/\epsilon)^{\frac1{n-1}}$. Consequently, if $\epsilon \to 0$ then $h(x) \to \infty$. 


The question now is, at which point(s) $x\_{\rm min} \in \mathbb H\_c^n$ is this minimum located, and what is the value $h\_{\rm min} = h(x\_{\rm min})$? 


Any minimum of $h$ on $\mathbb H\_c^n$ must also be a stationary point of $h$ on $\mathbb H\_c^n$ (i.e., the gradient of $h$ in $H\_c^n$ must vanish at that point). I believe that for any given $n$ there is only one stationary point of $h$ on $\mathbb H\_c^n$, which then also must be its (unique) minimum. Unfortunately, I cannot prove this yet, but I can prove the existence of a remarkably simple stationary point:



> 
> *Proposition 2*: 
>  Let us choose $c = \frac14$ in the constraint (2). The point $x$ with components 
>  \begin{align\*}
> x\_1 = x\_n = \frac12 , \quad x\_2 = \cdots = x\_{n-1} = 1
> \tag\*{(3)}
> \end{align\*}
>  is a stationary point of $h$ on $\mathbb H\_{1/4}^n$, with corresponding stationary value $h(x) = \frac{n^2}4$. 
> 
> 
> 


This extremely simple result suggests that there may be a simple reason why it is true, but so far I have only a (relatively) complicated proof which I won't write down here. 


**Question**: 


1. Why is the form (3) of the stationary point this simple?
2. Is (3) the *only* stationary point (and thus the global minimum) of $h$ under the constraint (2) with $c = \frac14$?



Surprisingly simple minimum of a rational function on $\mathbb R_+^n$
I'm not sure about stationary points, but the global minimum is certainly there. Let's do it for $n=5$. Write
$$
f(x)=1+\frac{x\_2}{2x\_1}+\frac{x\_2}{2x\_1}+\frac{x\_2x\_4}{2x\_1x\_3}+\frac{x\_2x\_4}{2x\_1x\_3}\,,
\\
g(x)=\frac 12+\frac 12+\frac{x\_3}{2x\_2}+\frac{x\_3}{2x\_2}+\frac{x\_3x\_5}{x\_2x\_4}
$$
Now use Cauchy-Schwarz ($(\sum a)(\sum b)\ge (\sum\sqrt{ab})^2$) honestly coupling the terms as presented and then apply AM-GM.


