# Surprisingly simple minimum of a rational function on $\mathbb R_+^n$

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 $c = \frac14$?

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.