Inequalities 2

Applying $\text AM \geq GM$ to calculate the maximum value of $x^{3}y^{3}$ ,

$\frac{x+x+x+y+y+y}{6} \geq (x^{3}y^{3})^{\frac{1}{6}}$

$x^{3}y^{3} \leq 1$

$\boxed{xy \leq 1}$

Now, $(x+y)^{3}=2^{3}$

$x^{3}+ y^{3}+3xy(x+y)=8$

$x^{3}+ y^{3}+3xy(2)=8$

To maximize $x^{3}+ y^{3}$ , putting $xy=1$

$x^{3}+ y^{3}+6 \leq 8$

$\boxed{x^{3}+ y^{3} \leq 2}$

Combining both,

$\boxed{(x^{3}+ y^{3})x^{3}y^{3} \leq 2}$