$\frac{a^{2}}{b+c}+\frac{b+c}{4}\geq 2\sqrt{\frac{a^{2}}{b+c}.\frac{b+c}{4}}=a$(1)$\frac{b^{2}}{c+a}+\frac{c+a}{4}\geq b$(2)
$\frac{c^{2}}{a+b}+\frac{a+b}{4}\geq c $(3)
(1)+(2)+(3)$\Rightarrow P\geq \frac{a+b+c}{2}\geq\frac{3\sqrt[3]{abc} }{2}=\frac{3}{2}$(abc=1)
Dau"="$\Leftrightarrow$a=b=c=1