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Given $x, y in mathbb{R}^+$ and $ x^3 + y^3 = x-y$. Prove $x^2 + 4y^2 < 1$

Mathematics Asked by user9026 on September 14, 2020

For $x, y in mathbb{R}^+$ , I am given that $ x^3 + y^3 = x-y$. I have to prove that $x^2 + 4y^2 < 1$. Now I have $ y + y^3 = x(1-x^2) $. Since $y + y^3 > 0 $ and $ x > 0$, we have $1 – x^2 > 0$. Which means that $ 0 < x < 1$. Also, since $x^3 + y^3 > 0$, we have $x – y > 0$. So, $x > y$. So, we get $0 < y < x < 1$. Using AM-GM inequality, I also get

$$ (x^3 + y^3 + 1) geqslant 3xy $$

What else can be dome here ?

Thanks

One Answer

We need to prove that $$x^2+4y^2<frac{x^3+y^3}{x-y}$$ or since $x-y>0$, $$y(5y^2-4xy+x^2)>0,$$ which is obvious because $$5y^2-4xy+x^2=y^2+(2y-x)^2>0.$$

Correct answer by Michael Rozenberg on September 14, 2020

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