# a i s The arithmetic mean of two positive numbers b and c . G_1 and G_2 are the geometric mean between the same positive numbers b and c so prove that G_1^3+G_2^3=2abc ?

Aug 9, 2018

#### Explanation:

Here,

color(blue)("a is the AM of b and c " =>(b+c)/2=a=>b+c=2ato(1)

Now, color(red)(G_1 and G_2 " are the GM between b and c."

So, $b , {G}_{1} , {G}_{2} , c \text{ are in GP}$

Let $r \text{ be the common ratio and b is the first term.}$

So, $b , b r , b {r}^{2} , b {r}^{3} \text{ are in GP}$

i.e. color(blue)(G_1=br ,G_2=br^2 ,c=br^3......to(2)

Let us take LHS.

$L H S = {\left({G}_{1}\right)}^{3} + {\left({G}_{2}\right)}^{3}$

$L H S = {\left(b r\right)}^{3} + {\left(b {r}^{2}\right)}^{3}$

$L H S = {b}^{3} {r}^{3} + {b}^{3} {r}^{6}$

$L H S = {b}^{2} \textcolor{b l u e}{\left(b {r}^{3}\right)} + b {\left(\textcolor{b l u e}{b {r}^{3}}\right)}^{2}$

LHS=b^2color(blue)((c))+bcolor(blue)((c)^2....to["using " (2) ]

$L H S = b c \left\{\textcolor{b l u e}{b + c}\right\}$

$L H S = b c \left\{\textcolor{b l u e}{2 a}\right\} \ldots \ldots \ldots \ldots \ldots \to \left[\text{using } \left(1\right)\right]$

$L H S = 2 a b c$

$L H S = R H S$