orl wrote: Let $ x$ and $ y$ be positive real numbers with $ x^3 + y^3 = x - y.$ Prove that \[ x^2 + 4y^2 < 1. \] Since $ x,y \geq 0$ then $ x - y = x^3 + y^3 \geq 0$ or $ x \geq y \geq 0$ Consider $ y = 0$ separately. If $ y > 0$ then \[ x^2 + 4y^2 < 1 \] Equivalent \[ (x - y)(x^2 + 4y^2) \leq x^3 + y^3 \]or \[ y(y^2 + (x - 2y)^2) > 0 \]

\[ x^2 + y^2<x^2 + 4y^2 < 1.\] The following inequality ( All-Russian Olympiad 1981) is also true. Let $ x$ and $ y$ be positive real numbers with $ x^3 + y^3 = x - y.$ Prove that \[ x^2 + y^2 < 1.\] Strengthen of CGMO 2005 Inspired by CGMO 2005 here

Another way: Note that $x>y$ Let $x=ky$. Then equation becomes $(k^3+1)y^3=(k-1)y$ So $y^2=\frac{k-1}{k^3+1}$ and now $x^2+4y^2=(k^2+4)y^2=\frac{(k^3-k^2+4k-4)}{k^3+1}$ which can be easily verified as $<1$

sqing wrote: \[ x^2 + y^2<x^2 + 4y^2 < 1.\]The following inequality ( All-Russian Olympiad 1981) is also true. Let $ x$ and $ y$ be positive real numbers with $ x^3 + y^3 = x - y.$ Prove that \[ x^2 + y^2 < 1.\]Strengthen of CGMO 2005 Inspired by CGMO 2005 here http://www.artofproblemsolving.com/community/c6h1388075p7728332

sqing wrote: The following inequality ( All-Russian Olympiad 1981) is also true. Let $ x$ and $ y$ be positive real numbers with $ x^3 + y^3 = x - y.$ Prove that \[ x^2 + y^2 < 1.\] All Soviet Union MO 1981 Let $x,y>0 ,x^3+y^3=x-y$ and $x^2+4y^2<1.$ Prove that $$x^2+5y^2 > 4xy$$

24. Let $x$ and $y$ be positive reals satisfying $x^3+y^3=x-y$. Find the maximum real $k$ such that for all $a<k$ have $x^2+ay^2<1$. $\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 3\sqrt[3]{3}\qquad \mathrm{(C) \ } 2+2\sqrt{2} \qquad \mathrm{(D) \ } 2\sqrt{6} \qquad \mathrm{(E) \ } 5$ 2015 Mock AMC10 Let $ a>b>0$ and $a^5 + b^5 =a-b.$ For $k<\frac{37}{3},$ prove that$$a^4 +kb^4 <1.$$

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Gibbenergy wrote: orl wrote: Let $ x$ and $ y$ be positive real numbers with $ x^3 + y^3 = x - y.$ Prove that \[ x^2 + 4y^2 < 1. \] Since $ x,y \geq 0$ then $ x - y = x^3 + y^3 \geq 0$ or $ x \geq y \geq 0$ Consider $ y = 0$ separately. If $ y > 0$ then \[ x^2 + 4y^2 < 1 \]Equivalent \[ (x - y)(x^2 + 4y^2) \leq x^3 + y^3 \]or \[ y(y^2 + (x - 2y)^2) > 0 \] Could you explain your motivation of your $(x-y)(x^2+4y^2) \leq x^3+y^3$?

Suan_16 wrote: Gibbenergy wrote: orl wrote: Let $ x$ and $ y$ be positive real numbers with $ x^3 + y^3 = x - y.$ Prove that \[ x^2 + 4y^2 < 1. \] Since $ x,y \geq 0$ then $ x - y = x^3 + y^3 \geq 0$ or $ x \geq y \geq 0$ Consider $ y = 0$ separately. If $ y > 0$ then \[ x^2 + 4y^2 < 1 \]Equivalent \[ (x - y)(x^2 + 4y^2) \leq x^3 + y^3 \]or \[ y(y^2 + (x - 2y)^2) > 0 \] Could you explain your motivation of your $(x-y)(x^2+4y^2) \leq x^3+y^3$? $$(x-y)(x^2+4y^2) < x^3+y^3 \iff x^2+5y^2 > 4xy$$