Question

If a and b are two positive quantities whose sum is $\lambda$, then the minimum value of $\sqrt{(1+\frac{1}{a})(1+\frac{1}{b})}$ is

• A. $\lambda -\frac{1}{\lambda }$
• B. $\lambda -\frac{2}{\lambda }$
• C. $1+\frac{1}{\lambda }$
• D. $1+\frac{2}{\lambda }$

Answer 1

The correct option is D $1+\frac{2}{\lambda }$

$E^2=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=1+\frac{a+b+1}{ab}=1+\frac{\lambda +1}{ab}$
Above will be minimum when ab is maximum.
Now we know that if sum of two quantities is constant then their product is maximum when the quantities are equal.
$\therefore a+b=\lambda \Rightarrow a=b=\frac{\lambda }{2}$
$\therefore E^2=\frac{{\lambda }^2+4\lambda +4}{{\lambda }^2}={\left(\frac{\lambda +2}{\lambda }\right)}^2$
$\therefore E=\frac{\lambda +2}{\lambda }=1+\frac{2}{\lambda }\Rightarrow \left(d\right)$