MHB Inequality challenge for positive real numbers

AI Thread Summary
The discussion centers on proving the inequality 2(√2 - 1)a² - b² < 1 under the condition that a and b are positive real numbers satisfying a³ + b³ = a - b. Participants agree that equality will never hold in this scenario. There is a consensus that the original problem statement needed correction to reflect that it should not imply a tight inequality. The conversation highlights the importance of accurately framing mathematical problems. Overall, the focus is on establishing the inequality while clarifying the conditions under which it holds.
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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