Meaners gonna mean mean mean.

Setting $m=\frac{a+b}{2}$ , $d=\frac{a-b}{2}=m-b$ and $g=\sqrt{ab}$ it is easy to show that $m^{2}=g^{2}+d^{2}$ , since the arithmetic mean, $m$ and geometric mean, $g$ are whole numbers $(d,g,m)$ is a Pythagorean triple.

So we must have $d=kx, g=ky,m=kz$ where $k$ is a whole-number and $(x,y,z)$ is a co-prime Pythagorean triple.

Since the harmonic mean $h$ satisfies $h=\frac{g^2}{m}$ we can see that $h=\frac{ky^{2}}{z}$ , which is a whole number only if $k$ is a multiple of $z$ . This means that $m=r z^{2}$ where $r$ is a whole number ( we also get that ( $g=r x z, h=r y^{2}, a=r z(z+x), b=r z (z-x)$ ). The smallest value of $m$ will be from the smallest Pythagorean triple (3,4,5) and setting $r=1$ , giving the smallest arithmetic mean is 25.

In general all values of $a$ and $b$ can be obtained from Pythagorean Triples.

Let $gcd(a, b)=m$ . Then, there exist distinct positive integers $x, y$ such that $a=xm$ and $b=ym$ with $gcd(x, y)=1$ .

Analyzing the harmonic mean, we notice that: $\dfrac{2ab}{a+b}=\dfrac{2m^2xy}{m(x+y)}=\dfrac{2mxy}{x+y}$ . However, since $gcd(x+y, x)=gcd(x+y, y)=1$ $\Rightarrow{x+y|2m}\Rightarrow{2m=(x+y)g}$ , for some positive integer $g$ .

From the geometric mean, we can deduce: $\sqrt{ab}=m\sqrt{xy}$ . But, $x, y$ are coprime, which implies both $x$ and $y$ are perfect squares.

Since we are looking for the minimum value, we will start checking individual, small cases for $x$ and $y$ , which will lead us to the minimum of $\dfrac{a+b}{2}$ .

If $x=1$ and $y=4\Rightarrow{g=2}$ . Then, $m=5\wedge{a=5, b=20\Rightarrow{\dfrac{a+b}{2}\notin{\mathbb{Z}^+}}}$ .

$g=4\Rightarrow{m=10\Rightarrow{a=10\wedge{b=40}}}$ $\Rightarrow{\dfrac{a+b}{2}=25}$ .

If $g=1\Rightarrow{x=1\wedge{y=9\Rightarrow{m=5}}}$ . This way, $a=5\wedge{b=45\Rightarrow{\dfrac{a+b}{2}=25}}$ . We conclude then that the minimum value for $\dfrac{a+b}{2}$ is $25$ .

Checking all square numbers such that they can be factored into a distinct $a$ and $b$ and that $a$ and $b$ are either both even or both odd, otherwise it would not be even and the arithmetic mean would not be a natural number. Now the harmonic mean is $\frac2{\frac1a+\frac1b}=\frac2{\frac{a+b}{ab}}=\frac{2ab}{a+b}$ therefore $a+b|2ab$ so we must check this condition. We can skip prime squares as they can only be written a single way. After a tedious search we arrive at $15^2=225=45\times5$ . These two numbers satisfy our conditions and we simply compute the arithmetic mean which is $25$ .

Use WolframAlpha Diophantine to find (40,10) and (45,5) as the pair with minimum mean of 25.