Question

If $a$ and $b$ are arbitary positive real numbers, then the least possible value of $\frac{6a}{5b}+\frac{10b}{3a}$ is

• A. $4$
• B. $\frac{6}{5}$
• C. $\frac{10}{3}$
• D. $\frac{68}{15}$

Answer 1

The correct option is A $4$

$AM\ge GM\frac{\frac{6a}{5b}+\frac{10b}{3a}}{2}\ge {(\frac{6a}{5b}.\frac{10b}{3a})}^{\frac{1}{2}}\Rightarrow \frac{6a}{5b}+\frac{10b}{3a}\ge 2\cdot 4^{\frac{1}{2}}$
$\Rightarrow \frac{6a}{5b}+\frac{10b}{3a}\ge 4$