Ta có: $P+11=2+\frac{3\left( b+c \right)}{2\text{a}}+1+\frac{4\text{a}+3c}{3b}+\frac{12\left( b-c \right)}{2\text{a}+3c}=\left( 4\text{a}+3b+3c \right)\left( \frac{1}{2\text{a}}+\frac{1}{3b}+\frac{4}{2\text{a}+3c} \right)$

Theo bất đẳng thức Cauchy - Schwarz ta có

                        $P+11\ge \left( 4\text{a}+3b+3c \right)\frac{16}{4\text{a}+3b+3c}=16\Rightarrow P\ge 15$

Đẳng thức xảy ra khi $b=c=\frac{2}{3}a$