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MathematicsSequence & SeriesMeans / inequalityMedium2 minQB

MathematicsMediumsingle choice

If $a, b, c$ are positive real numbers such that $a b^{2} c^{3} = 64$ , then the minimum value of $\frac{1}{a} + \frac{2}{b} + \frac{3}{c}$ is equal to

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Solution:

Take the six numbers $\frac{1}{a}, \frac{1}{b}, \frac{1}{b}, \frac{1}{c}, \frac{1}{c}, \frac{1}{c}$ . A.M. $\geq$ G.M.: $\frac{\frac{1}{a} + \frac{2}{b} + \frac{3}{c}}{6} \geq (\frac{1}{a b ^{2} c ^{3}})^{1/6} = (\frac{1}{64})^{1/6} = \frac{1}{2} .$

So $\frac{1}{a} + \frac{2}{b} + \frac{3}{c} \geq 3$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Means / inequality

⏱ 2mℹ️ Source: QB

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