Answer: $\mathbf B$
Explanation:
Let $a + b = \lambda$
$\therefore b = \lambda - a$
Let $f = \sqrt{({1 + \frac{1}{a}})(1 + \frac{1}{b})}$
Let $g = f^2 =({1 + \frac{1}{a}})(1 + \frac{1}{b})$
Taking derivative of $g$ with respect to $a$ and making it equal to $0$, we get:
$g' = \frac{-1}{a^2}(1+\frac{1}{\lambda-a}) + (1+\frac{1}{a})(\frac{1}{(\lambda-a)^2}) = 0$
Simplifying this we get:
$(\lambda-2a)(\lambda+1) = 0, \;or \; a = \frac{\lambda}{2}$
Again on differentiating $g$ with respect to a, we get:
$g^{''} = (\frac{2}{a^3})(1+\frac{1}{\lambda-a}) - \frac{1}{a^2}\frac{1}{(\lambda-a)^2} + (1+\frac{1}{a})(\frac{2}{(\lambda-a)^3})-\frac{1}{a^2}(\frac{1}{(\lambda-a)^2})$
At $a = \frac{\lambda}{2}$ the value of $g^{''}>0$
$\Rightarrow $ It's a point of minima.
$\therefore$ Minimum value $= \frac{\lambda+2}{\lambda}$
$\therefore \mathbf B$ is the correct option.
