Explanation

(b)
Since, $\frac{(1 + a_{1})}{2} \geq \sqrt{1 \bullet a_{1}} = \sqrt{a_{1}}$
$\frac{(1 + a_{2})}{2} \geq \sqrt{1 \bullet a_{2}} = \sqrt{a_{2}}$
$\ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$
$\frac{(1 + a_{n})}{2} \geq \sqrt{1 \bullet a_{n}} = \sqrt{a_{n}}$
$\Rightarrow \frac{1}{2^{n}}\left( 1 + a_{1} \right)\left( 1 + a_{2} \right)\ldots\left( 1 + a_{n} \right) \geq \sqrt{a_{1}a_{2}\ldots a_{n}} = 1$
$\Rightarrow \left( 1 + a_{1} \right)\left( 1 + a_{2} \right)\ldots(1 + a_{n}) \geq 2^{n}$