Proving x^4 + 1 > x^9 + x with x <= 1

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SUMMARY

The inequality x^4 + 1 > x^9 + x is proven to hold true under the condition that x ≤ 1. The transformation of the inequality into the form (x - 1)(x^8 + x^7 + x^6 + x^5 + x^4 + 1) < 0 reveals that the product is zero at x = 1 and positive for x > 1. The analysis confirms that for values of x between 0 and 1, the inequality holds, while it also holds true at x = 0. Further exploration is needed for values of x less than 0.

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Homework Statement



Let x[tex]\in[/tex][tex]\textbf{R}[/tex]. Prove that if x4+1 > x9+x , then x[tex]\leq[/tex]1

Homework Equations



[above]

The Attempt at a Solution



Sort of clueless on this one. My intuition tells me I should algebraically rearrange the equation such that 1 is on one side of the inequality. I also considered factoring out an x.
 
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Maybe this will help you get started.
x4 + 1 > x9 + x
<==> x9 - x4 + x - 1 < 0
<==> (x - 1)(x8 + x7 + x6 +x5 + x4 + 1) < 0

I got this factorization by using synthetic division.

If x = 1, the product is zero, so the inequality doesn't hold. If x > 1, both factors are positive, so the inequality again doesn't hold.

Can you say anything about the product if 0 < x < 1?

The inequality holds if x = 0. Can you say something about the product if x < 0?
 

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