Question

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
(A) 4
(B) \(\frac{6}{5}\)
(C) \(\frac{10}{3}\)
(D) \(\frac{68}{15}\)

Answer 1

Ans: (A)
Hint \(: \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \sqrt{\frac{6 a}{5 b}} \times \frac{10 b}{3 a}, \quad \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \times 2 \geq 4\)

Answer 2

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