Integers are mean

The AM, GM, and HM form a geometric sequence. Therefore, they can be expressed as $km^2$ , $kmn$ , and $kn^2$ respectively, where $k$ , $m$ , and $n$ are all integers.

Then, using the definitions of the harmonic mean: $\frac{2xy}{x+y}=kn^2$

Manipulating with the equation: $2xy-kn^2x-kn^2y=0 \\ 2x(2y-kn^2)-2kn^2y=0 \\ 2x(2y-kn^2)-2kn^2y+k^2n^4=k^2n^4 \\ (2x-kn^2)(2y-kn^2)=k^2n^4$

Splitting the product gives: $2x-kn^2=kn^2\left(\frac ab\right) \\ 2y-kn^2=kn^2\left(\frac ba\right)$ where $a$ and $b$ are coprime.

Changing the subject: $x=\frac{kn^2(a+b)}{2b} \\ y=\frac{kn^2(a+b)}{2a}$

Since $xy$ is a perfect square, $ab$ is also a perfect square. Then, we can split it into $\frac{cp}d$ and $\frac{dp}c$ where $ab=p^2$ , and $c$ and $d$ are coprime.

Then: $x=\frac{kn^2\left(\frac{cp}{d}+\frac{dp}{c}\right)}{\frac{2dp}{c}} = \frac{kn^2(c^2+d^2)}{2d^2} \\ y=\frac{kn^2\left(\frac{cp}{d}+\frac{dp}{c}\right)}{\frac{2cp}{d}} = \frac{kn^2(c^2+d^2)}{2c^2}$

Now, $c$ and $d$ are free variables, while $k$ and $n$ are semi-free variables.

$\mbox{AM}=\frac{kn^2(c^2+d^2)^2}{4c^2d^2} \\ \mbox{GM}=\frac{kn^2(c^2+d^2)}{2cd} \\ \mbox{HM}=kn^2$

Since $(c^2+d^2)$ is not divisible by $c^2d^2$ because $c$ and $d$ are coprime, $kn^2$ must be divisible by $c^2d^2$ . Let the quotient be $q$ : $x = \frac{qc^2(c^2+d^2)}2 \\ y = \frac{qd^2(c^2+d^2)}2 \\ \mbox{AM} = \frac{q(c^2+d^2)^2}4 \\ \mbox{GM} = \frac{qcd(c^2+d^2)}2 \\ \mbox{HM} = qc^2d^2$

Apparently the solution is obtained when $c$ and $d$ are as low as possible, i.e. $2$ and $1$ . This yields $x=40$ , $y=10$ , $\mbox{AM}=25$ , $\mbox{GM}=20$ , $\mbox{HM}=16$ , and the required sum would be $111$ .

However notice that the solution might be smaller when $c^2+d^2$ is divisible by $2$ . Then, we shall try $c=3$ and $d=1$ , which gives us $x=45$ , $y=5$ , $\mbox{AM}=25$ , $\mbox{GM}=15$ , $\mbox{HM}=9$ , and the required sum would be $\mbox{99}$ , which is even smaller than $111$ .

When $c$ and $d$ get bigger, so does the sum. Therefore $99$ is the minimum.