4. Let $x$ and $y$ be positive numbers whose sum is 2. Find the maximum value of the expression $x^{2} y^{2}\left(x^{2}+y^{2}\right)$.
Solution 1. The idea is suggested by the formula for $(x+y)^{4}$. Since

$(x+y)^{4}-8 x y\left(x^{2}+y^{2}\right)=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}=(x-y)^{4} \geqslant 0$, then $8 x y\left(x^{2}+y^{2}\right) \leqslant(x+y)^{4}$. It is clear that $4 x y \leqslant(x+y)^{2}$. Therefore,

$$
32 x^{2} y^{2}\left(x^{2}+y^{2}\right) \leqslant(x+y)^{6}
$$

Since $x+y=2$, then $x^{2} y^{2}\left(x^{2}+y^{2}\right) \leqslant 2$. It is clear that when $x=y=1$ we get equality.

Answer: 2.