a+b+c=1
=>c=-a-b+1; b=-a-c+1; a=-b-c+1
ab+c
=ab-a-b+1
=a(b-1)-(b-1)
=(a-1)(b-1)
\(\left(a+b\right)^2=\left(1-c\right)^2=\left(c-1\right)^2\)
bc+a
=bc-b-c+1
=b(c-1)-(c-1)
=(c-1)(b-1)
\(\left(b+c\right)^2=\left(1-a\right)^2=\left(a-1\right)^2\)
ca+b
=ca-c-a+1
=c(a-1)-(a-1)
=(a-1)(c-1)
\(\left(c+a\right)^2=\left(1-b\right)^2=\left(b-1\right)^2\)
\(P=\frac{ab+c}{\left(a+b\right)^2}\cdot\frac{bc+a}{\left(b+c\right)^2}\cdot\frac{ca+b}{\left(c+a\right)^2}\)
\(=\frac{\left(a-1\right)\left(b-1\right)}{\left(c-1\right)^2}\cdot\frac{\left(b-1\right)\left(c-1\right)}{\left(a-1\right)^2}\cdot\frac{\left(a-1\right)\cdot\left(c-1\right)}{\left(b-1\right)^2}=1\)
