Question

If $a, b, c > 0$ and if $abc = 1$, 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

Answer: (D)

$[6,\infty,)$

Answer 2

Since $A.M.\ge G.M.$ $\Rightarrow \frac{a+b+c}{3} 3\sqrt{abc}$ $\Rightarrow \frac{a+b+c}{3} \ge\left(1\right)^{\frac{1}{3}}$ $\quad\left(\because abc=1\right)$ $\Rightarrow a+b+c \ge 3\quad...\left(1\right)$ Also $G.M. \ge H.M.$ $\Rightarrow \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ $\Rightarrow\left(1\right)^{\frac{1}{3}}\ge \frac{3abc}{ab+bc+ca}$ $\Rightarrow ab+bc+ca \ge3\left(1\right)$ $\Rightarrow ab+bc+ca \ge3$ $\qquad\quad...\left(2\right)$ Adding $\left(1\right)$ and $\left(2\right)$, we get $a+b+c+ab+bc+ca\ge6$ $\therefore a+b+c+ab+bc+ca$ lies in $[6, \infty)$