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- Feb 14, 2012

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Prove $x+x^9+x^{25}<1+x^4+x^{16}+x^{36}$ for $x>0$.

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

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Prove $x+x^9+x^{25}<1+x^4+x^{16}+x^{36}$ for $x>0$.

- Aug 30, 2012

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Let's see if my head is fully back in the game.

Let's look at \(\displaystyle f(x) = x^{36} - x^{25} + x^{16} - x^9 + x^4 - x + 1\)

For x = 0, f(0) = 1.

For 0 < x < 1:

\(\displaystyle x^{36} - (x^{25} - x^{16} ) - (x^9 - x^4) - (x - 1) > 0\) because each term inside the parentheses are negative.

For 1 < x

\(\displaystyle (x^{36} - x^{25}) + (x^{16} - x^9) + (x^4 - x) + 1 > 0\) because every term inside the parentheses are positive.

Therefore f(x) has no real zeros on 0 < x.

\(\displaystyle x^{36} - x^{25} + x^{16} - x^9 + x^4 - x + 1 > 0 \implies x^{36} + x^{16} + x^4 + 1 > x^{25} + x^9 + x\) on 0 < x.

-Dan

For x = 0, f(0) = 1.

For 0 < x < 1:

\(\displaystyle x^{36} - (x^{25} - x^{16} ) - (x^9 - x^4) - (x - 1) > 0\) because each term inside the parentheses are negative.

For 1 < x

\(\displaystyle (x^{36} - x^{25}) + (x^{16} - x^9) + (x^4 - x) + 1 > 0\) because every term inside the parentheses are positive.

Therefore f(x) has no real zeros on 0 < x.

\(\displaystyle x^{36} - x^{25} + x^{16} - x^9 + x^4 - x + 1 > 0 \implies x^{36} + x^{16} + x^4 + 1 > x^{25} + x^9 + x\) on 0 < x.

-Dan

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