A good inequality problem!

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

This is a wonderful problem which can be solved using some clever manipulations and without much of the inequality results.

Do note that your task is to find the most elegant, short and understandable solution to this problem.

Note by Sathvik Acharya
11 months, 2 weeks ago

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  Easy Math Editor

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Replace \( a \) by \( x^5 - x^3 + x \) in the inequality to get the equivalent:

\[ x^6 - 2x^5 + 2x^3 -2x + 1 \geq 0 \]

The polynomial on the left-hand side is seen to have the same coefficients when \( x \) is replaced by \( \frac{1}{x} \). This means we can factorise it with not too much difficulty as:

\[ x^6 - 2x^5 + 2x^3 -2x + 1 = (x - 1)^2 \cdot ((x^2 - 1)^2 + x^2) \geq 0 \]

which is true. Equality at x = 1 follows.

Ameya Daigavane - 11 months ago

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Nice one!

Sathvik Acharya - 11 months ago

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