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If \(x^5 - x^3 + x = a\) for real \(a\), then prove that \(x^6 \geq 2a - 1\).

Note by Ankit Kumar Jain 1 year, 5 months ago

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\(\displaystyle \begin{align} x^6+1 &= (x^2+1)(x^4-x^2+1) \\ &= \dfrac{x^2+1}{x}(x^5-x^3+x) \\ &= \dfrac{x^2+1}{x}a \\ &\ge 2a \end{align}\)

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Thanks!! @Aditya Narayan Sharma

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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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TopNewest\(\displaystyle \begin{align} x^6+1 &= (x^2+1)(x^4-x^2+1) \\ &= \dfrac{x^2+1}{x}(x^5-x^3+x) \\ &= \dfrac{x^2+1}{x}a \\ &\ge 2a \end{align}\)

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Thanks!! @Aditya Narayan Sharma

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