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

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: $(1+\alpha)(1+\beta)(1+\gamma)(1+\delta)$ $=1+\alpha+\beta+\gamma+\delta+\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta+\alpha \beta \gamma+\beta \gamma \delta+\alpha \gamma \delta+\alpha \beta \delta+\alpha \beta \gamma \delta$ $=1+(\alpha+\beta+\gamma+\delta)+(\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta)+(\alpha \beta \gamma+\beta \gamma \delta+\alpha \gamma \delta+\alpha \beta \delta)+\alpha \beta \gamma \delta$ $=1+(\alpha+\beta+\gamma+\delta)+\left(\alpha \beta+\alpha \gamma+\alpha \delta+\frac{1}{\alpha \delta}+\frac{1}{\alpha \gamma}+\frac{1}{\alpha \beta}\right)+\left(\frac{1}{\delta}+\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)+1$ $\ldots \ldots\{\because \alpha \beta \gamma \delta=1\}$ $= 2+\left(\alpha+\frac{1}{\alpha}\right)+\left(\beta+\frac{1}{\beta}\right)+\left(\gamma+\frac{1}{\gamma}\right)+\left(\delta+\frac{1}{\delta}\right)+\left(\alpha \beta+\frac{1}{\alpha \beta}\right)+\left(\alpha \gamma+\frac{1}{\alpha \gamma}\right)+\left(\alpha \delta+\frac{1}{\alpha \delta}\right) $ $\geq 2+2+2+2+2+2+2+2 $ $\left\{\because x +\frac{1}{ x } \geq 2 \right.$ for $\left. x \geq 0\right\}$ $\geq 16$ Hence, the least value of $(1+\alpha)(1+\beta)(1+\gamma)(1+\delta)$ is $16$ .