\(P=\frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}=\frac{bc}{a\left(a+b+c\right)+bc}+\frac{ca}{b\left(a+b+c\right)+ca}+\frac{ab}{c\left(a+b+c\right)+ab}\)
\(=\frac{bc}{\left(a+b\right)\left(a+c\right)}+\frac{ca}{\left(a+b\right)\left(b+c\right)}+\frac{ab}{\left(a+c\right)\left(b+c\right)}\)
\(=\frac{bc\left(b+c\right)+ca\left(c+a\right)+ab\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) \(=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)-2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1-\frac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
+ Ta có : \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge3\sqrt[3]{abc}\cdot3\sqrt[2]{a^2b^2c^2}-abc=8abc\)
\(\Rightarrow P\ge1-\frac{2abc}{8abc}=1-\frac{1}{4}=\frac{3}{4}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{3}\)
