1. Determine all real numbers $p$ such that for any positive numbers $x, y$ the inequality

$$
\frac{x^{3}+p y^{3}}{x+y} \geqq x y
$$

holds.
1. The inequality is clearly equivalent to the relation for positive $x, y$

$$
y^{2}(p y-x)-x^{2}(y-x) \geqq 0 .
$$

If $p \geqq 1$, we get

$$
y^{2}(p y-x)-x^{2}(y-x) \geqq y^{2}(y-x)-x^{2}(y-x)=(x+y)(y-x)^{2} \geqq 0 \text {. }
$$

However, if $p<1$, the given inequality does not hold, for example, for $x=y=1$.

Conclusion: The given inequality is satisfied for any two positive numbers $x, y$, if and only if $p \geqq 1$.