$x^3+y^3+3x^2+3y^2+4x+4y+4=0$ $(x; y>0)$

$\Leftrightarrow$ $(x^3+y^3)+3(x^2+y^2)+4(x+y)+4=0$

$\Leftrightarrow$ $(x+y)^3-3xy(x+y)+3(x+y)^2-6xy+4(x+y)+4=0$

$\Leftrightarrow$ $[(x+y)^3+2(x+y)^2+(x+y)^2+4(x+y)+4]-[3xy(x+y)+6xy]=0$

$\Leftrightarrow$ $[(x+y)^2(x+y+2)+(x+y+2)^2]-3xy(x+y+2)=0$

$\Leftrightarrow$ $(x+y+2)[(x+y)^2+x+y+2]-3xy(x+y+2)=0$

$\Leftrightarrow$ $(x+y+2)[(x+y)^2+x+y+2-3xy]=0$

$\Leftrightarrow$ $(x+y+2)(x^2+2xy+y^2+x+y+2-3xy)=0$

$\Leftrightarrow$ $(x+y+2)[(x^2-2xy+y^2)+(x^2+2x+1)+(y^2+2y+1)+2]:2=0$

$\Leftrightarrow$ $(x+y+2)[(x-y)^2+(x+1)^2+(y+1)^2+2]=0$

$\Leftrightarrow$ $x+y+2=0$

$\Leftrightarrow$ $x+y=-2$

Ta có: `M=1/x+1/y=(x+y)/(xy)=(-2)/(xy)`

Vì $4xy$ $\le$ $(x+y)^2$

$\Leftrightarrow$ $4xy$ $\le$ $(-2)^2$

$\Leftrightarrow$ $4xy$ $\le$ $4$

$\Leftrightarrow$ $xy$ $\le$ $1$

$\Rightarrow$ `1/(xy)` $\ge$ $1$

$\Leftrightarrow$ `(-2)/(xy)` $\le$ $-2$ ($xy>0$)

$\Leftrightarrow$ $M$ $\le$ $-2$

Dấu $=$ xảy ra khi $x=y=-1$ (loại vì $x; y>0$)

Vậy không có GTLN của $M$