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  4. If a, b >0, then the minimum value of the expression ((1+a+a2)(1+b+b2)/a b) is equal to

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Q. If $a, b >0$, then the minimum value of the expression $\frac{\left(1+a+a^2\right)\left(1+b+b^2\right)}{a b}$ is equal to

Complex Numbers and Quadratic Equations

Solution:

$E=\left(a+\frac{1}{a}+1\right)\left(b+\frac{1}{b}+1\right) \geq 9$
$\therefore E _{\min }=9$ at $a =1= b$.