Question

For all $x,y\in +R$, the value of $\frac{(1+x+x^2)(1+y+y^2)}{xy}$, is

• A. $\le 9$
• B. $<9$
• C. $\ge 9$
• D. $>9$

Answer 1

Answer: (C)

$\ge 9$

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

$⟹\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

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

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

$⟹\frac{1+x+x^2}{3}\ge x$

$⟹\frac{1+x+x^2}{x}\ge 3$ ------(1)

similarly,

$⟹\frac{1+y+y^2}{y}\ge 3$------(2)

multiplying (1) and (2) we get

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

$⟹\frac{(1+x+x^2)(1+y+y^2)}{xy}\ge 9$