MHB Inequality Challenge: Prove $x$ for $x>0$

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The inequality to prove is \(x + x^9 + x^{25} < 1 + x^4 + x^{16} + x^{36}\) for \(x > 0\). Participants discuss various approaches to proving this inequality, including analyzing the behavior of both sides as \(x\) approaches 0 and infinity. Some suggest using calculus to find critical points and evaluate the function's growth rates. Others propose testing specific values of \(x\) to verify the inequality's validity. The discussion emphasizes the need for a rigorous mathematical proof to confirm the inequality holds for all positive \(x\).
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Prove $x+x^9+x^{25}<1+x^4+x^{16}+x^{36}$ for $x>0$.
 
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Let's see if my head is fully back in the game.

Let's look at [math]f(x) = x^{36} - x^{25} + x^{16} - x^9 + x^4 - x + 1[/math]

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

For 0 < x < 1:
[math]x^{36} - (x^{25} - x^{16} ) - (x^9 - x^4) - (x - 1) > 0[/math] because each term inside the parentheses are negative.

For 1 < x
[math](x^{36} - x^{25}) + (x^{16} - x^9) + (x^4 - x) + 1 > 0[/math] because every term inside the parentheses are positive.

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

[math]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[/math] on 0 < x.

-Dan
 
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