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If $a$ and $b$ are arbitrary positive real numbers, then the least possible value of $\frac{6 a}{5 b}+\frac{10 b}{3 a}$ is
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Verified Answer
The correct answer is:
4
Hint:
$\frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \sqrt{\frac{6 a}{5 b} \times \frac{10 b}{3 a}}, \quad \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \times 2 \geq 4$
$\frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \sqrt{\frac{6 a}{5 b} \times \frac{10 b}{3 a}}, \quad \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \times 2 \geq 4$
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