\(P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\) \(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
Áp dụng Cauchy-Schwarz dạng phân thức:
\(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{9}{2\left(a+b+c\right)}\)
\(\Leftrightarrow P+3\ge\frac{9}{2}\Rightarrow P\ge\frac{3}{2}\)
\(''=''\Leftrightarrow a=b=c\)