Question

If three positive real numbers a, b, c are in A.P., with abc=4, then the minimum value of b is

• A. $\frac{1}{2}$
• B. 2
• C. ${\left(4\right)}^{\frac{1}{3}}$
• D. 3

Answer 1

The correct option is C

${\left(4\right)}^{\frac{1}{3}}$

Using AM ≥ GM

$\frac{a+b+c}{3}$ ≥ ${\left(abc\right)}^{\frac{1}{3}}$

$\frac{3b}{3}$ ≥ ${\left(abc\right)}^{\frac{1}{3}}$ (since 2b = a + c)

b ≥ ${\left(4\right)}^{\frac{1}{3}}$