Question

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

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

Answer 1

The correct option is D

$\ge 9$

Find the value of $\frac{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{y}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{)}}{\mathbf{x}\mathbf{y}}$:

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

We know that $AM\ge GM$

$$\begin{gathered} \Rightarrow \frac{(a+b+c)}{3}\ge {\left(abc\right)}^{1/3} \\ \Rightarrow \frac{1}{3}(1+x+x^2)\ge {\left(x^3\right)}^{1/3} \\ \Rightarrow \frac{(1+x+x^2)}{x}\ge 3....\dots \left(i\right) \end{gathered}$$

Similarly

$\frac{(1+y+y^2)}{y}\ge 3...\dots \left(ii\right)$

Multiply (i) and (ii). we get

$$\begin{gathered} \Rightarrow \frac{(1+x+x^2)(1+y+y^2)}{xy}\ge 3\times 3 \\ \Rightarrow \frac{(1+x+x^2)(1+y+y^2)}{xy}\ge 9 \end{gathered}$$

Hence the correct option is D.