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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 6 months ago

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\(\displaystyle \begin{align} x^6+1 &= (x^2+1)(x^4-x^2+1) \\ &= \dfrac{x^2+1}{x}(x^5-x^3+x) \\ &= \dfrac{x^2+1}{x}a \\ &\ge 2a \end{align}\) – Aditya Narayan Sharma · 3 months, 1 week ago

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@Aditya Narayan Sharma – Thanks!! @Aditya Narayan Sharma – Ankit Kumar Jain · 3 months ago

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TopNewest\(\displaystyle \begin{align} x^6+1 &= (x^2+1)(x^4-x^2+1) \\ &= \dfrac{x^2+1}{x}(x^5-x^3+x) \\ &= \dfrac{x^2+1}{x}a \\ &\ge 2a \end{align}\) – Aditya Narayan Sharma · 3 months, 1 week ago

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@Aditya Narayan Sharma – Ankit Kumar Jain · 3 months ago

Thanks!!Log in to reply