relation between am, gm and hm 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?
 A: 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}
