System

[ins-]

Solve in real numbers the system:

$\sqrt{1+(x+y)^2}=-y^6+2x^2y^3+4x^4$
$\sqrt{2x^2y^2-x^4y^4} \ge 4x^2y^3+5x^3$

[mihailm]

Даже если писать по английски, то нужны попытки решения.

[ins-]

I managed to prove $x^4 \ge x^3$, but I cannot continue from here.
To get this conclusion I used the following chain of inequalities: $\sqrt{2x^2y^2-x^4y^4} \le 1 \le \sqrt{1+(x+y)^2}$

[ProPupil]

First of all, you should find the domain of the system. The first square root is always defined while the other is not, so first of all you have

$2x^2y^2-x^4y^4 \geq 0$ which leads us to $x^2y^2 \leq 2$

Then, arithmetical square root is always positive which leads us to

$-y^6+2x^2y^3+4x^4 \geq 0$ and $4x^2y^3+5x^3 \geq 0$

Try seeing whether those domains intersect or not.

[ins-]

$4x^2y^3+5x^3 \ge 0$ I'm not sure this is correct, because it is possible to have $\sqrt{2x^2y^2 - x^4 y^4} \ge 0 \ge 4x^2y^3+5x^3$.
The other idea I have is: $ 1 \le \sqrt{1+(x+y)^2} = -y^6 + 2x^2y^3+4x^4 \le 5x^4 $

(Оффтоп)

I suppose it can be useful information in solving the problem. Its source is Bulgarian Math Olympiad - 1979 -Regional Round: http://imomath.com/othercomp/Bul/BulMO379.pdf I have a book, dedicated to that competition and it is written the only solution of this system is: (1,-1). The interesting moment is - this problem was left with no solution. It leads me to the conclusion - it should not be a difficult problem, but for some reason I'm out of ideas.

[ins-]

http://artofproblemsolving.com/communit ... 86p2762286 I found a very similar system here. It can be useful in the solving process of the original system I posted.

[ins-]

http://www.wolframalpha.com/input/?i=%5 ... %5E%7B3%7D it seems there is some mistake related to this problem or the software.

[ins-]

Based on previous post I suppose the correct system should be:

Solve in real numbers the system:
$\sqrt{1+(x+y)^2}=-y^6+2x^2y^3+4x^4$
$\sqrt{2x^2y^2-x^4y^4} \ge 4x^2y^3+5x^4$

This system is not hard and even I can solve it