WBJEE · Maths · Sequences and Series
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 & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
Hint: \(\frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \sqrt{\frac{6 a}{5 b} \times \frac{10 b}{3 a}}, \quad \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \times 2 \geq 4\)
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