Concyclicboy wrote: The arithmetic, geometric and harmonic mean of two distinct positive integers are different numbers. Find the smallest possible value for the arithmetic mean. Huhh ! choosing as distinct positive integers $1,2$, we verify easily that arithmetic, geometric and harmonic mean are different numbers and the arithmetic mean is $\frac 32$ and is obviously the smallest possible. Maybe I misunderstood something ?

I think the means must be different integer numbers.

Let our positive integers be $a \neq b$. Let $d = \gcd(a,b)$, thus $a=dm$, $b=dn$, with $\gcd(m,n)=1$. The harmonic mean $\frac{2}{\frac{1}{a}+\frac{1}{b}}$ is an integer, therefore $a+b \mid 2ab$, which means $d(m+n) \mid 2d^2mn \Rightarrow m+n \mid 2dmn \Rightarrow m+n \mid 2d$, because $m+n$ and $mn$ are relatively prime. The arithmetic mean is an integer, thus $2 \mid d(m+n)$. The geometric mean is an integer, hence $d^2mn$ is a perfect square $\Rightarrow mn$ is a perfect square $\Rightarrow m$ and $n$ are both perfect squares, because they are relatively prime. We want to find the minimum value of $\frac{d(m+n)}{2}$ under these conditions. If $m+n$ is odd, we get $m+n \mid d$ and $2 \mid d$, thus $d \geq 2(m+n)$. But $m+n \geq 1+4 = 5$, thus $\frac{d(m+n)}{2} \geq \frac{10 \times 5}{2} = 25$. If $m+n$ is even, we get $d \geq \frac{m+n}{2}$, since $m+n \geq 1+9 = 10$, again we get $\frac{d(m+n)}{2} \geq \frac{5 \times 10}{2} = 25$. So the answer is $25$. Examples attaining this minimum value are $(5,45)$, $(10,40)$.

Also posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=378710

Here's another solution. We can write $a=k+i,b=k-i$ and $(k+i)(k-i)=m^2$, ie. $k^2=m^2+i^2$ for $k,m,i$ positive integers. We want to minimize $k$. HM being an integer implies $k\mid k^2-i^2$, meaning $k\mid i^2$ and $k\mid m^2$. So we can write $i^2=kc, m^2=kd$ for some positive integers $c,d$. We can see that $k=c+d$. So we've translated the problem to finding $c,d$ such that $c(c+d)$ and $d(c+d)$ are perfect squares and $c+d$ is minimal. Note that if $c$ and $d$ are not coprime, then letting $(c,d)=x$ we can find another pair $c'=\frac{c}{x},d'=\frac{d}{x}$ that also works, and which makes their sum smaller. So if $c$ and $d$ are minimal, they must be coprime. This means $c,d,c+d$ are all perfect squares; letting $c=x^2,d=y^2$, we have $x^2+y^2=z^2$ for coprime $x, y, z$. It's easy to check that the lowest value of $z^2$ in such a triad is $25$; namely, we have $3^2+4^2=5^2$. Thus the minimum value for $z^2=c+d=k$ is $25$.

Let $a, b$ be distinct positive integers such that their arithmetic, geometric, and harmonic means are distinct integers. If $d=\gcd(a, b)$ and $a=d\alpha, b=d\beta$, then the following are integers: \[ \frac{d}2\cdot(\alpha+\beta), d\sqrt{\alpha\beta}, \frac{2d\alpha\beta}{\alpha+\beta}. \]Since $\gcd(\alpha, \beta)=1,$ this means $\alpha+\beta\mid2d$ and there are positive integers $x\ne y$ such that $a=dx^2, b=dy^2$. Thus, the following are integers: \[ \frac{d}2(x^2+y^2), dxy, \frac{2d}{x^2+y^2}. \]Since $x\ne y$, we know $x^2+y^2\ge 5$. Also, if $x^2+y^2>5$ then it is at least $10$. In the second case $d\ge 5$. In the first case, $5\mid d$. However, since at least one of $x^2+y^2$ is even, $d\ge 10$. In the first case, \[ \frac{a+b}2\ge\frac{10+40}2=25. \]In the second, \[ \frac{a+b}2\ge\frac{5+45}2=25. \]So, the answer is $25$ which can be achieved with either $5, 45$ or $10, 40$. $\blacksquare$