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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

  1. A 4
  2. B \(\frac{6}{5}\)
  3. C \(\frac{10}{3}\)
  4. D \(\frac{68}{15}\)
Verified Solution

Answer & Solution

Correct Answer

(A) 4

Step-by-step Solution

Detailed explanation

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