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

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SUMMARY

The inequality challenge presented is to prove that for all positive values of \(x\), the expression \(x + x^9 + x^{25} < 1 + x^4 + x^{16} + x^{36}\) holds true. Participants in the discussion utilized algebraic manipulation and properties of exponents to analyze the inequality. The consensus is that the inequality can be verified through specific substitutions and bounding techniques, confirming its validity for \(x > 0\).

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  • Understanding of algebraic inequalities
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  • Basic knowledge of calculus for bounding functions
  • Experience with mathematical proofs
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  • Study algebraic manipulation techniques for inequalities
  • Learn about bounding functions and their applications in proofs
  • Explore the use of calculus in proving inequalities
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Mathematics students, educators, and enthusiasts interested in advanced algebra and inequality proofs.

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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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