If a and b are arbitrary positive real number | vedclass

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(A) To find the least value of the expression $\frac{6a}{5b} + \frac{10b}{3a}$,we use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.For

Vedclass · Accepted Answer

$4$

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(A) To find the least value of the expression $\frac{6a}{5b} + \frac{10b}{3a}$,we use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.For any two positive real numbers $x$ and $y$,the inequality states that $\frac{x+y}{2} \geq \sqrt{xy}$,or $x+y \geq 2\sqrt{xy}$.Let $x = \frac{6a}{5b}$ and $y = \frac{10b}{3a}$.Then,$\frac{6a}{5b} + \frac{10b}{3a} \geq 2 \sqrt{\frac{6a}{5b} 	imes \frac{10b}{3a}}$.Simplifying the expression inside the square root:$\frac{6a}{5b} 	imes \frac{10b}{3a} = \frac{6 	imes 10}{5 	imes 3} 	imes \frac{a}{a} 	imes \frac{b}{b} = \frac{60}{15} = 4$.Therefore,$\frac{6a}{5b} + \frac{10b}{3a} \geq 2 \sqrt{4} = 2 	imes 2 = 4$.The least possible value is $4$.

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

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