Question

For all positive value of $x$ and $y$, 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

The correct option is C $\ge 9$

Solution:

To find: $\frac{(1+x+x^2)(1+y+y^2)}{xy}=?$

We know that,

$A.M.>G.M.$

$\frac{1+x+x^2}{xy}=\frac{1}{xy}+\frac{1}{y}+\frac{x}{y}$

$\therefore \frac{\frac{1}{xy}+\frac{1}{y}+\frac{x}{y}}{3}\ge {(\frac{1}{xy}.\frac{1}{y}.\frac{x}{y})}^{1/3}$

$\therefore \frac{1}{xy}+\frac{1}{y}+\frac{x}{y}\ge \frac{3}{y}.............................\left(i\right)$

Similarly,

$\frac{1+y+y^2}{xy}=\frac{1}{xy}+\frac{1}{x}+\frac{y}{x}$

$\therefore \frac{\frac{1}{xy}+\frac{1}{x}+\frac{y}{x}}{3}\ge {(\frac{1}{xy}.\frac{1}{x}.\frac{y}{x})}^{1/3}$

$\therefore \frac{1}{xy}+\frac{1}{x}+\frac{y}{x}\ge \frac{3}{x}.............................\left(ii\right)$

On multiplying eqn (i) and (ii), we get

$(\frac{1}{xy}+\frac{1}{y}+\frac{x}{y})(\frac{1}{xy}+\frac{1}{x}+\frac{y}{x})\ge \frac{9}{xy}$

$⟹\frac{(1+x+x^2)(1+y+y^2)}{(xy{)}^2}\ge \frac{9}{xy}$

$\therefore \frac{(1+x+x^2)(1+y+y^2)}{xy}\ge 9$

Hence, C is the correct option.