For the positive real numbers $x, y$, we know that

$$
x^{5}+5 x^{3} y+5 x^{2} y^{2}+5 x y^{3}+y^{5}=1
$$

What positive values can $x+y$ take?
$$
x^{5}+5 x^{3} y+5 x^{2} y^{2}+5 x y^{3}+y^{5}=1
$$

Solution I. Let the two-variable function in the problem be denoted by $f(x, y)$. It is easily seen that

$$
f(x, y)=x^{5}+y^{5}+\frac{(x+y)^{5}-x^{5}-y^{5}}{x+y}
$$

From this, it is clear that if $x+y=1$, then the equality (1) holds. We will show that if $x+y>0$ and $f(x, y)=1$, then neither $x+y>1$ nor $x+y<1$ can hold. 

If $x+y>1$, then since $x+y>0$, we have $x^{5}+y^{5}>0$, so if $x+y>1$, then

$$
\frac{x^{5}+y^{5}}{x+y} < x^{5}+y^{5}+(x+y)^{4}-\left(x^{5}+y^{5}\right)
$$

which simplifies to

$$
1>(x+y)^{4}
$$

This is impossible since $x+y>1$.

Similarly, if $x+y<1$, and $x+y>0$, then the second factor in (2) is positive, so from (1) it indeed follows that $x+y=1$.

Remark. The condition that $x$ and $y$ are positive (or the weaker condition $x+y>0$) cannot be omitted from the problem's conditions. For a fixed $y$, the second factor in (2) is a fifth-degree polynomial in $x$, which has a real root, and this root is usually not $1-y$.