relation between am, gm and hm

Question

if $A, G$ and $H$ are respectively the arithmetic mean, geometric mean and the harmonic mean of $n$ positive numbers, what are the conditions under which the equation $$G^2=A\times H$$ holds?

Answer 1

The condition

\begin{align} G^2&=A\times H \tag{1}\label{1} \end{align}

for $n$ positive numbers $x_1,\dots,x_n$ always holds trivially, when they all are the same, hence in this case $A=H=G=x_k$.

For $n=3$ there are plenty of suitable distinct triplets, for example, $(2, 3+\sqrt5, 3-\sqrt5)$.

Let's try to construct a list of $n$ positive numbers for which condition \eqref{1} holds. Starting with arbitrary $n-1$ positive numbers, define \begin{align} a&=\sum_{k=1}^{n-1}x_k ,\\ g&=\prod_{k=1}^{n-1}x_k ,\\ h&=\sum_{k=1}^{n-1}\frac1{x_k} , \end{align}

let's find out what value of $x_n=v$ we should add to the list to get \eqref{1}. Value $v$ completes the list, so we can calculate

\begin{align} A&=\tfrac1n(a+v) ,\\ G&=(g\,v)^{1/n} ,\\ H&=\frac{n}{h+\tfrac1v} , \end{align} and \eqref{1} becomes

\begin{align} (g\,v)^{2/n}&= \frac{(a+v)v}{h\,v+1} \end{align}

or in a polynomial form,

\begin{align} g^2 (h\,v+1)^n-(a+v)^n\,v^{n-2}&=0 \tag{2}\label{2} . \end{align}

Example for $n=5$. Let's try the four numbers be of the form $x_k=\tfrac1k$. Then

\begin{align} a&=\sum_{k=1}^{4}\tfrac1k = \tfrac{25}{12} ,\\ g&=\prod_{k=1}^{4}\tfrac1k = \tfrac1{24} ,\\ h&=\sum_{k=1}^{4} k= 10 . \end{align}

Polynomial \eqref{2} then becomes

\begin{align} \tfrac1{576}(10v+1)^5-(\tfrac{25}{12}+v)^5 v^3 , \end{align}

which, luckily, has a nice rational root $v=\tfrac34$, so the list of five numbers is now complete: $X_5=(1,\tfrac12,\tfrac13,\tfrac14,\tfrac34)$.

Checking \eqref{1} for $X_5$:

\begin{align} A&=\tfrac{17}{30},\quad G=\tfrac12,\quad H=\tfrac{15}{34} ,\\ G^2&=A\cdot H=\tfrac14 . \end{align}