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I have to prove the following inequality: $$ x^2+4y^2<1$$ with this constrain $$x-y=x^3+y^3$$ and where $x$ and $y$ are positive real numbers.

From the constrain we have also $0<x<1$.

I can't put the constrain in the inequality in some useful form.

Any help is appreciated, thanks.

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  • $\begingroup$ $x^3+y^3-x+y=0\Rightarrow x(x^2-1)+y(y^2+1)=0$ Because $x$ and $y$ are positives we have also $y(y^2+1)>0$ and so $x(x^2-1)$ must be negative $\Rightarrow x^2-1 <0$ $\endgroup$ – Alex Apr 18 at 9:06
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Let's prove that for x>y>0 the following holds $$x^2 +4y^2< \frac{x^3+y^3}{x-y}$$

$$ (x^2 + 4y^2)(x-y) < x^3 + y^3 $$

$$ x^3 -x^2y +4y^2x-4y^3<x^3+y^3$$

Now cancel out the x^3 and divide by y.

$$ -x^2 +4xy -4y^2 < y^2$$

$$4xy< x^2 + 5y^2$$

and that is true by AM-GM: $x^2 +5y^2 >=2\sqrt{5}xy>4xy$

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Since $x,y> 0$ then $x-y=x^3+y^3> 0$ or $x-y> 0$.

So we have $x^2+4y^2<1\iff (x-y)(x^2+4y^2)<x^3+y^3\iff y(y^2+(x-2y)^2)>0$(TRUE).

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Let $y=mx$

$$x-y=x^3+y^3\implies x^2=\dfrac{1-m}{1+m^3},y^2=?$$

$$x^2+4y^2-1=\dfrac{1-m}{1+m^3}+\dfrac{4m^2(1-m)}{1+m^3}-1=\dfrac{1-m+4m^2-4m^3}{1+m^3}-1$$

$$=\dfrac{-m(5m^2-4m+1)}{1+m^3}=\dfrac{-m(m^2+(2m-1)^2)}{1+m^3}<0$$

as $m,2m-1$ can not be $=0$ together

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