The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.)

$$AM=GM+2$$

$$GM=HM+1.6$$

Since $$GM^2=AM\cdot HM,$$

$$GM^2=(GM+2)(GM-1.6)$$

$$GM^2=GM^2+0.4GM-3.2$$

$$GM=8$$

$$AM=10$$

$$\sqrt{ab}=8, \frac{a+b}{2}=10$$

$$ab=64, a+b = 20$$

The numbers are $16$ and $4$.

I don't know why I do this, but here is the general case.

Suppose $am = gm+u$ and $gm = hm+v$ with $u, v \ne 0$ and $u \ne v$.

Since $hm \le gm \le am $, $u \ge 0$ and $v \ge 0$.

Since $gm^2 = am\cdot hm$, $gm^2 =(gm+u)(gm-v) =gm^2+gm(u-v)-uv $ so $gm(u-v) =uv$. Therefore $u > v$ and $gm =\dfrac{uv}{u-v} $.

Then $am =gm+u =\dfrac{uv}{u-v}+u =\dfrac{uv+u(u-v)}{u-v}+u =\dfrac{u^2}{u-v} $ and $hm =gm-v =\dfrac{uv}{u-v}-v =\dfrac{uv-v(u-v)}{u-v} =\dfrac{v^2}{u-v} $.

If there are only two values, $a$ and $b$, then $\dfrac{a+b}{2} =\dfrac{u^2}{u-v} $ and $\sqrt{ab} =\dfrac{uv}{u-v} $ so $b =\dfrac{u^2v^2}{a(u-v)^2} $ and $\dfrac{u^2}{u-v} =\dfrac{a+\dfrac{u^2v^2}{a(u-v)^2}}{2} =\dfrac{a^2(u-v)^2+u^2v^2}{2a(u-v)^2} $ or $2au^2(u-v) =a^2(u-v)^2+u^2v^2 $.

Solving $a^2(u-v)^2-2u^2(u-v)a+u^2v^2 =0$,

$\begin{array}\\ a &=\dfrac{2u^2(u-v)\pm\sqrt{4u^4(u-v)^2-4(u-v)^2u^2v^2}}{2(u-v)^2}\\ &=\dfrac{u^2(u-v)\pm u(u-v)\sqrt{u^2-v^2}}{(u-v)^2}\\ &=\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\ &=u\dfrac{u\pm \sqrt{u^2-v^2}}{u-v}\\ \text{and}\\ b &=\dfrac{2u^2}{u-v}-a\\ &=\dfrac{2u^2}{u-v}-\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\ &=\dfrac{2u^2-u^2\mp u\sqrt{u^2-v^2}}{u-v}\\ &=\dfrac{u^2\mp u\sqrt{u^2-v^2}}{u-v}\\ \end{array} $

For this case, $u=2$ and $v=1.6$.

$gm =\dfrac{2\cdot 1.6}{2-1.6} =\dfrac{3.2}{.4} =8 $.

To get $a$ and $b$, $\sqrt{u^2-v^2} =\sqrt{4-2.56} =\sqrt{1.44} =1.2 $ so $a =2\dfrac{2\pm 1.2}{.4} =2\dfrac{3.2, .8}{.4} =2(8, 2) =(16, 4) $ and $b =(4, 16) $.