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Prove that the reciprocal of the sum of the reciprocals of 2 naturals number is always less than the sum of the natural numbers.

Note by Aditya Dhawan 2 years, 2 months ago

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Let the two numbers be \(a\) and \(b\).

Since the numbers are positive, we can use AM-HM inequality.

\[\frac{a + b}{2} \geq \frac{2}{\frac{1}{a} + \frac{1}{b}}\]

\[\implies \frac{1}{\frac{1}{a} + \frac{1}{b}}\leq \frac{a + b}{4} < a + b~~~~~ _{\square}\] – Pranshu Gaba · 2 years, 2 months ago

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TopNewestLet the two numbers be \(a\) and \(b\).

Since the numbers are positive, we can use AM-HM inequality.

\[\frac{a + b}{2} \geq \frac{2}{\frac{1}{a} + \frac{1}{b}}\]

\[\implies \frac{1}{\frac{1}{a} + \frac{1}{b}}\leq \frac{a + b}{4} < a + b~~~~~ _{\square}\] – Pranshu Gaba · 2 years, 2 months ago

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