Đặt:$\left\{ {\begin{array}{*{20}{c}}
{a + b + c = x}&{}&{}\\
{4a + b + c = 2y}&{}&{}\\
{16b + c + a = 6z}&{}&{}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{a = \frac{{2y - x}}{3}}&{}&{}\\
{b = \frac{{6z - x}}{{15}}}&{}&{}\\
{c = \frac{{21x - 10y - 6z}}{{15}}}&{}&{}
\end{array}} \right.$

$ \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{a + b = \frac{{10y + 6z - 6x}}{{15}}}&{}&{}\\
{b + c = \frac{{4x - 2y}}{3}}&{}&{}\\
{c + a = \frac{{16x - 6z}}{{15}}}&{}&{}
\end{array}} \right.$

Khi đó ta có: $\left\{ {\begin{array}{*{20}{c}}
{\frac{{a + b}}{{a + b + c}} = \frac{{10y + 6z - 6x}}{{15x}} = \frac{{2y}}{{3x}} + \frac{{2z}}{{5x}} - \frac{2}{5}}&{}&{}\\
{\frac{{b + c}}{{4a + b + c}} = \frac{{4x - 2y}}{{6y}} = \frac{{2x}}{{3y}} - \frac{1}{3}}&{}&{}\\
{\frac{{c + a}}{{16b + c + a}} = \frac{{16x - 6z}}{{90z}} = \frac{{8x}}{{45z}} - \frac{1}{{15}}}&{}&{}
\end{array}} \right.$

Từ đó, ta viết lại thành: $P = \left( {\frac{{2x}}{{3y}} + \frac{{2y}}{{3x}}} \right) + \left( {\frac{{2z}}{{5x}} + \frac{{8x}}{{45z}}} \right) - \frac{4}{5} \ge \frac{4}{3} + \frac{8}{{15}} - \frac{4}{5} = \frac{{16}}{{15}}$

Vậy GTNN của $P$ là $\frac{16}{15}$