Minimum value#4

From above assumption,

$\begin{cases} 1\le x\le 2 \\ 1\le y\le 2 \end{cases}\Longrightarrow \begin{cases} \left( x-1 \right) \left( x-2 \right) ={ x }^{ 2 }-3x+2\ge 0 \\ \left( y-1 \right) \left( y-2 \right) ={ y }^{ 2 }-3y+2\ge 0 \end{cases}\Longrightarrow \begin{cases} { x }^{ 2 }\le 3x-2 \\ { y }^{ 2 }\le 3y-2 \end{cases}$

Hence,

$P\ge \frac { x+2y }{ 3x+3y+3 } +\frac { y+2x }{ 3y+3x+3 } +\frac { 1 }{ 4\left( x+y-1 \right) } =\frac { x+y }{ x+y+1 } +\frac { 1 }{ 4\left( x+y-1 \right) } =Q$

Let $u=x+y$ then $2\le u\le 4$ , and

$Q=f\left( u \right) =\frac { u }{ u+1 } +\frac { 1 }{ 4\left( u-1 \right) } .$

We now have

$f^{ ' }\left( u \right) =\frac { 1 }{ { \left( u+1 \right) }^{ 2 } } -\frac { 1 }{ 4{ \left( u-1 \right) }^{ 2 } } =\frac { \left( 3u-1 \right) \left( u-3 \right) }{ 4{ \left( u+1 \right) }^{ 2 }{ \left( u-1 \right) }^{ 2 } } .$

We notice that $f\left( u \right)$ reaches its minimum value when $u=3$ .

Therefore, $P\ge Q\ge f\left( 3 \right) =\frac { 7 }{ 8 } =0.875$

Finally, ${ P }_{ min }=\boxed { 0.875 }$ when $\left( x,y \right) =\left( 1,2 \right) ,\left( 2,1 \right) .$