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

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Homework Statement



Let x\in\textbf{R}. Prove that if x4+1 > x9+x , then x\leq1

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?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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