How find this value of $x,y$ let $x,y\in R$, such


  $$\begin{cases}
\sqrt{1+(x+y)^2}=-y^6+2x^2y^3+4x^4\\
\sqrt{2x^2y^2-x^4y^4}\ge 4x^2y^3+5x^3
\end{cases}$$
  find the value of  $x,y$.


My try:
since


  $$|xy|\sqrt{2-x^2y^2}\ge 4x^2y^3+5x^3$$


then I can't,it is said this problem is Bulgaria National Olympiad(1979),But I can't find it .Thank you
Surely there's a whole family of solutions to this, if all you know is $x,y\in\mathbb{R}$.  How about trying $y=0$?  Then the first equation becomes
$$
\sqrt{1+x^2}=4x^4
$$
and the second becomes
$$
0 \ge 5x^3,
$$
or $x\le 0$.  The first equation has a solution for negative $x$ (about $-0.7474$), so that's one answer.  Is there something special about the particular solution you need?  (Or does "R" mean something different in Bulgaria?)
Surely there's a whole family of solutions to this, if all you know is $x,y\in\mathbb{R}$.  How about trying $y=0$?  Then the first equation becomes
$$
\sqrt{1+x^2}=4x^4
$$
and the second becomes
$$
0 \ge 5x^3,
$$
or $x\le 0$.  The first equation has a solution for negative $x$ (about $-0.7474$), so that's one answer.  Is there something special about the particular solution you need?  (Or does "R" mean something different in Bulgaria?)