Inequality challenge for positive real numbers

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Discussion Overview

The discussion revolves around an inequality challenge involving two positive real numbers, \(a\) and \(b\), under the condition that \(a^3 + b^3 = a - b\). Participants are tasked with proving the inequality \(2(\sqrt{2}-1)a^2 - b^2 < 1\) or its variants.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant asserts that under the given condition, the inequality \(2(\sqrt{2}-1)a^2 - b^2 < 1\) should be proven.
  • Another participant suggests that the inequality should be \(2(\sqrt{2}-1)a^2 - b^2 \le 1\) and emphasizes that equality will never hold.
  • A third participant agrees with the second viewpoint, indicating that the original problem should not involve a tight inequality and expresses gratitude for the correction.

Areas of Agreement / Disagreement

Participants express disagreement regarding the form of the inequality, with some advocating for a strict inequality while others propose a non-strict version. There is no consensus on the correct formulation of the inequality.

Contextual Notes

The discussion highlights the potential for misunderstanding in the formulation of inequalities, particularly in the context of the given condition \(a^3 + b^3 = a - b\). The implications of the inequality's tightness remain unresolved.

anemone
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If $a$ and $b$ are two positive real, and that $a^3+b^3=a-b$, prove that $2\left(\sqrt{2}-1\right)a^2-b^2\lt 1$.
 
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anemone said:
If $a$ and $b$ are two positive real, and that $a^3+b^3=a-b$, prove that $2\left(\sqrt{2}-1\right)a^2-b^2\le 1$.
it should be:If $a$ and $b$ are two positive real, and that $a^3+b^3=a-b$, prove that $2\left(\sqrt{2}-1\right)a^2-b^2<1$.
equality will never hold
 
anemone said:
If $a$ and $b$ are two positive real, and that $a^3+b^3=a-b$, prove that $2\left(\sqrt{2}-1\right)a^2-b^2\le 1$.
$a^3+b^3=a-b>0 \,\,\therefore a>b$,
$a-a^3=b+b^3>0,\therefore a<1$
we have:$0<b<a<1$
and $2\left(\sqrt{2}-1\right)a^2-b^2<2(1.5-1)\times 1^2-0^2=1$
equality will never hold
 
Ah, Albert, you're absolutely right, the problem itself should not be a tight inequality. I will fix it now, and thanks for catching it!
 

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