The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) by $2$ and the GM exceeds the Harmonic Mean (H.M) by $1.6$. Find the numbers.
My Attempt:
Let the numbers be $a$ and $b$. Then,
$$A.M=\dfrac {a+b}{2}$$
$$G.M=\sqrt {ab}$$
$$H.M=\dfrac {2ab}{a+b}$$
According to question:
$$\dfrac {a+b}{2} =\sqrt {ab}+2$$
$$\dfrac {a+b}{2}-2=\sqrt {ab}$$
$$a+b-4=2\sqrt {ab}$$
Also,
$$\sqrt {ab}=\dfrac {2ab}{a+b} + 1.6$$
Then,
$$a+b-4=2(\dfrac {2ab}{a+b} + 1.6)$$
$$a+b-4=\dfrac {4ab+3.2(a+b)}{a+b}$$
$$(a+b-4)(a+b)=4ab+3.2(a+b)$$
$$(a+b)^2-4(a+b)=4ab+3.2a+3.2b$$
How do I solve further?

$$AM=GM+2$$
$$GM=HM+1.6$$
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 $4$.
