Junior Balkan Mathematical Olympiad Shortlist

Solution:

We have $(x+y)\left(x^{2}+y^{2}\right) \geq (x+y)\left(x^{3}+y^{3}\right) \geq \left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq (1+1)\left(x^{2}+y^{2}\right) \geq (x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y=1$.

So the greatest possible value of the product $x y$ is $1$.

By $AM-GM$ we have $x^{3}+y^{3} \geq \sqrt{x y} \cdot \left(x^{2}+y^{2}\right)$, hence $1 \geq \sqrt{x y}$ since $x^{2}+y^{2} \geq x^{3}+y^{3}$. Equality holds when $x=y=1$. So the greatest possible value of the product $x y$ is $1$.