Question

If a and b are positive real numbers such that $a^2{b}^3=\frac{4}{9},$ then least value of $a+b$ is

• A. $1$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{5}{3}$

Answer 1

Answer: (D)

$\frac{5}{3}$

$\frac{\frac{2a}{2}+\frac{3b}{3}}{2+3}\ge \left[\right(\frac{9}{2}{)}^2\left(\frac{b}{3}{)}^3\right]\frac{1}{5}$

$\frac{a+b}{5}\ge [\frac{a^2{b}^2}{108}{]}^{\frac{1}{5}}$

$\frac{a+b}{5}\ge (\frac{4}{9\times 108}{)}^{\frac{1}{5}}$

$a+b\ge \frac{5}{3}$