Question

If $a,b,c>0$and $abc=1$, then the value of $a+b+c+ab+bc+ca$ lies in the interval

• A. $(\infty ,-6)$
• B. $(-6,0)$
• C. $(0,6)$
• D. $[6,\infty )$

Answer 1

The correct option is D

$[6,\infty )$

Step1. Construction :

Since AM ≥ GM

$$\begin{gathered} \Rightarrow \frac{(a+b+c)}{3}\ge \sqrt[3]{abc} \\ \Rightarrow \frac{(a+b+c)}{3}\ge {\left(1\right)}^{1/3}\mathbf{}\mathbf{(}\mathbf{\because }\mathbf{}\mathit{a}\mathit{b}\mathit{c}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{)} \\ \Rightarrow a+b+c\ge 3\dots ....\left(1\right) \end{gathered}$$

And also GM ≥ HM

$$\begin{gathered} \Rightarrow \sqrt[3]{abc}\ge \frac{3}{\left[\right(1/a)+(1/b)+(1/c\left)\right]} \\ \Rightarrow {\left(1\right)}^{1/3}\ge \frac{3abc}{[ab+bc+ca]} \\ \Rightarrow ab+bc+ca\ge 3\times 1 \\ \Rightarrow ab+bc+ca\ge 3......\dots \left(2\right) \end{gathered}$$

Step2. Find the interval:

Adding (1) and (2), we get

$a+b+c+ab+bc+ca\ge 6$

$\therefore a+b+c+ab+bc+ca$ lies in $[6,\infty )$.

Hence, the correct option is D.