Ta có: \(a+b+c=1\Rightarrow\hept{\begin{cases}a=1-b-c\\b=1-a-c\\c=1-a-b\end{cases}}\)
\(\Rightarrow\left(ab+c\right)\left(bc+a\right)\left(ac+b\right)\)\(=\left(ab+1-a-b\right)\left(bc+1-b-c\right)\left(ac+1-a-c\right)\)
\(=\left[\left(ab-a\right)-\left(b-1\right)\right]\left[\left(bc-b\right)-\left(c-1\right)\right]\left[\left(ac-c\right)-\left(a-1\right)\right]\)
\(=\left[a\left(b-1\right)-\left(b-1\right)\right]\left[b\left(c-1\right)-\left(c-1\right)\right]\left[c\left(a-1\right)-\left(a-1\right)\right]\)
\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)\left(b-1\right)\left(a-1\right)\left(c-1\right)\)
\(=\left(a-1\right)^2\left(b-1\right)^2\left(c-1\right)^2\)
\(=\left(1-a\right)^2\left(1-b\right)^2\left(1-c\right)^2\)
