Có $a+9b\ge 6\sqrt{ab}$$8a+2c\ge8\sqrt{ca}$
$\Rightarrow P\ge\frac1{9(a+b+c)}+\frac{\sqrt{a+b+c}}{243}+\frac{\sqrt{a+b+c}}{243}+\frac{484\sqrt{a+b+c}}{243}\ge\frac1{27}+\frac{484}{81}=\frac{487}{81}$
Dấu bằng khi và chỉ khi $c=4a;a=9b;a+b+c=9$
$a=\frac{81}{46};b=\frac9{46};c=\frac{162}{23}$