<s>[INST] 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?
 [/INST] 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)
$.
</s>