$=>(x+1)^3+(y+1)^3+(x+1)+(y+1)=0$đặt $a=x+1;b=y+1$
$=>a^3+b^3+a+b=0=>\left\{ \begin{array}{l} a+b=0\\ a^2-ab+b^2+1=0 (VN) \end{array} \right.$
$=>x+y=-2$
$=>(x+y)^2=4\geq 4xy=>xy\leq 1$
$\frac{1}{x}+\frac{1}{y}=\frac{-2}{xy}\leq -2$
dấu $=$ xảy ra khi $x=y=-1$