# Algebraic Manipulation

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}

- 1 year, 2 months ago