<s>[INST] Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$ 
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$

From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2) = 4x^2y^2\\ 4x^2y^2 = (x + y)(1 + xy) \le \frac{[(x + 1)(y + 1)]^2}{4} \le \frac{(x + y + 2)^4}{4^3} \end{align*} \right.$$
$$\implies (x + y)^2 \le \frac{(x + y + 2)^4}{4^3} \implies |x + y| \le \left|\frac{x + y + 2}{2^3}\right|$$
I don't know what to do next. Please help me solve this problem.
 [/INST] 
My simple solution is $(\!x^{2}+ y^{2}- 2x^{2}y^{2}\!)+ \left (\!(\!x+ y\!)(\!1+ x^{2}y^{2}\!)\!\right )= (\!x+ y- 2xy\!)(\!3xy+ x+ y+ 1\!)$

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