Inequality challenge for positive real numbers

[anemone]

If $a$ and $b$ are two positive real, and that $a^3+b^3=a-b$, prove that $2\left(\sqrt{2}-1\right)a^2-b^2\lt 1$.

 

[Albert1]

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

 

[Albert1]

If $a$ and $b$ are two positive real, and that $a^3+b^3=a-b$, prove that $2\left(\sqrt{2}-1\right)a^2-b^2\le 1$.

Spoiler

$a^3+b^3=a-b>0 \,\,\therefore a>b$,
$a-a^3=b+b^3>0,\therefore a<1$
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

 

[anemone]

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!