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.

  • $\begingroup$ $x^3+y^3-x+y=0\Rightarrow x(x^2-1)+y(y^2+1)=0$ 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

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: $x^2 +5y^2 >=2\sqrt{5}xy>4xy$


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).


Let $y=mx$

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



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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.