1.If a and b are arbitrary positive real numbers, then the least possible value of 6a5b+10b3a\frac{6a}{5b} + \frac{10b}{3a}5b6a+3a10b isa.4b.65\frac{6}{5}56c.103\frac{10}{3}310d.6815\frac{68}{15}1568Login to continueOnly logged in users canattempt or see the solution.
