Question

Let $x$ be the arithmetic mean and $y,z$ be the two geometric means between any two positive numbers. Then $\frac{y^3+z^3}{xyz}$ = __________.

Answer 1

Given that $x$ is the A.M. between $a$ and $b$ and $y$, and $z$ are the G.M. between $a$ and $b$ where $a$ and $b$ are positive. Then $a,x,b$ are in A.P. So,

$x=\frac{a+b}{2}$

$a,y,z,b$ are in G.P. So,

$y=ar$ and $z=a{r}^2$, $b=zr\Rightarrow b=a{r}^3$
$\Rightarrow r=(\frac{b}{a}{)}^{\frac{1}{3}}$
Also, $yz=ab$,

Now, $\frac{y^3+z^3}{xyz}=\frac{a^3{r}^3+a^3{r}^6}{ab(\frac{a+b}{2})}$

$=\frac{a^3\times \frac{b}{a}+a^3\times \frac{b^2}{a^2}}{ab(\frac{a+b}{2})}$

$=\frac{2(a^{2}b+a{b}^2)}{a^{2}b+a{b}^2}$

$\therefore \frac{y^3+z^3}{xyz}=2$