Question

Let $\alpha ,\beta ,\gamma$ and $\delta$ are four positive real number such that their product is unity, then the least value of $(1+\alpha )(1+\beta )(1+\gamma )(1+\delta )$ is:

• A. $6$
• B. $16$
• C. $0$
• D. $32$

Answer 1

The correct option is B $16$

It is given that the product of $\alpha ,\beta ,\gamma$ and $\delta$ is unity that is $\alpha \beta \gamma \delta =1$.

Now consider $(1+\alpha )(1+\beta )(1+\gamma )(1+\delta )$ as follows:

$(1+\alpha )(1+\beta )(1+\gamma )(1+\delta )=1+\alpha +\beta +\gamma +\delta +\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta +\alpha \beta \gamma +\beta \gamma \delta +\alpha \gamma \delta +\alpha \beta \delta +\alpha \beta \gamma \delta =1+(\alpha +\beta +\gamma +\delta )+(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta )+(\alpha \beta \gamma +\beta \gamma \delta +\alpha \gamma \delta +\alpha \beta \delta )+\alpha \beta \gamma \delta$

$=1+(\alpha +\beta +\gamma +\delta )+(\alpha \beta +\alpha \gamma +\alpha \delta +\frac{1}{\alpha \delta }+\frac{1}{\alpha \gamma }+\frac{1}{\alpha \beta })+(\frac{1}{\delta }+\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })+1$ ....... $\{\because \alpha \beta \gamma \delta =1\}$

$=2+(\alpha +\frac{1}{\alpha })+(\beta +\frac{1}{\beta })+(\gamma +\frac{1}{\gamma })+(\delta +\frac{1}{\delta })+(\alpha \beta +\frac{1}{\alpha \beta })+(\alpha \gamma +\frac{1}{\alpha \gamma })+(\alpha \delta +\frac{1}{\alpha \delta })\ge 2+2+2+2+2+2+2+2\{\because x+\frac{1}{x}\ge 2forx\ge 0\}\ge 16$

Hence, the least value of $(1+\alpha )(1+\beta )(1+\gamma )(1+\delta )$ is $16$.