Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean …

Question

Let $G$ be the geometric mean of two positive numbers $a$ and $b$, and $M$ be the arithmetic mean of $\dfrac1a$ and $\dfrac1b$. If $\dfrac1M:G$ is $4:5$, then $a:b$ can be:

Accepted Answer

$G=\sqrt{ab}$, $M=\dfrac{\frac1a+\frac1b}{2}=\dfrac{a+b}{2ab}$, so $\dfrac1M=\dfrac{2ab}{a+b}$.

$$\frac{2ab}{(a+b)\sqrt{ab}}=\frac45 \Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac54 \Rightarrow \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{9}{1}$$
(by componendo--dividendo), i.e. $\left(\dfrac{\sqrt a+\sqrt b}{\sqrt b-\sqrt a}\right)^2=9 \Rightarrow \dfrac{\sqrt b+\sqrt a}{\sqrt b-\sqrt a}=3$.

Componendo--dividendo again: $\dfrac{2\sqrt b}{2\sqrt a}=\dfrac{4}{2}=2 \Rightarrow \dfrac ba=4 \Rightarrow a:b=1:4.$

Solutions · 1