I think the proof is immediate. Assume w.l.o.g. $x\le y$. We have by Chebyshev's Inequality that $x^3+y^3=x\cdot x^2+y\cdot y^2\ge y(x^2)+x(y^2)=xy(x+y)=p$. There is equality when $x=y=\displaystyle (\frac{p}{2})^{\frac{1}{3}}$.

José wrote: Let $p$ be a positive real number given. Find the minimun vale of $x^{3}+y^{3}$, knowing that $x$ and $y$ are positive real numbers such that $xy(x+y)=p$. It's easy to prove that: Let x,y >0, then we have: x^3+y^3>=xy(x+y), Prof: x^3+y^3=(x+y)(x^2+y^2-xy) From: x^2+y^2>=2xy then (x^2+y^2-xy)>=xy. From that it's proved. Quote:

This can be done using Schur's ineq, which states that: For any non-negative reals $x,y,z$ and any positive integer $t$ it holds that: $$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y) \ge 0$$Where equality holds when $x=y=z$ or when any $2$ of them are equal and the last one is $0$. Proof of the main problem: i claim that the minimun is $p$ and its reached when $x=y=\sqrt[3]{\frac{p}{2}}$ Now its enough to show that $x^3+y^3 \ge xy(x+y)$ which is equivant to $x(x)(x-y)+y(y)(y-x) \ge 0$ but this holds by Schur ineq when $z=0$ and $t=1$ hence the we are done