Question

If $a,b,c$ are three distinct positive real number , then the least value of
$\frac{(1+a+a^2)(1+b+b^2)(1+c+c^2)}{abc}$, is

• A. $3$
• B. $9$
• C. $27$
• D. None of the above.

Answer 1

The correct option is C $27$

since, $A.M.\ge G.M.$

$⟹\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

let $a=1,b=a$ and $c=a^2$

$⟹\frac{1+a+a^2}{3}\ge \sqrt[3]{31\ast a\ast a^2}$

$⟹\frac{1+a+a^2}{3}\ge a$

$⟹\frac{1+a+a^2}{a}\ge 3$ ------(1)

similarly,

$⟹\frac{1+b+b^2}{b}\ge 3$------(2)

and

$⟹\frac{1+c+c^2}{c}\ge 3$------(3)

multiplying (1), (2) and (3) we get

$⟹\left(\frac{1+a+a^2}{a}\right)\left(\frac{1+b+b^2}{b}\right)\left(\frac{1+c+c^2}{c}\right)\ge 3\ast 3\ast 3$

$⟹\frac{(1+a+a^2)(1+b+b^2)(1+c+c^2)}{abc}\ge 27$