Question

Let a, b and c be positive real numbers such that $a+b+c=6$ Then range of $a{b}^2{c}^3$ is

• A. (0,∞)
• B. (0, 1)
• C. (0, 108]
• D. (6, 108]

Answer 1

The correct option is C (0, 108]

$\frac{a+\frac{b}{2}.2+\frac{c}{3}}{6}\ge {\left(a\right(\frac{b}{2}{)}^2\left(\frac{c}{3}{)}^3\right)}^{1/3}$

$1\ge {\left(\frac{a{b}^2{c}^3}{2^{2}33}\right)}^{1/3}$

$a{b}^2{c}^3\le 108$

$(0,108]$