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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, 8 months ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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

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

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

\[\implies \frac{1}{\frac{1}{a} + \frac{1}{b}}\leq \frac{a + b}{4} < a + b~~~~~ _{\square}\]

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

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

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

\[\implies \frac{1}{\frac{1}{a} + \frac{1}{b}}\leq \frac{a + b}{4} < a + b~~~~~ _{\square}\]

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