Vì a + c = 2016 -> a = 2016 - [ b + c] ; b = 2016 - [ a + c] ; c = 2016 - [ a - b]
Ta có: S = a/ b + c + b/ a + c + c/a + b
S = 2016 - [ b + c] + 2016 - [ a + c] + 2016 - [ a + b]
S = 2016/ b + c - 1 + 2016/a + c - 1 + 2016/a + b
S = 2016.[ 1/b + c + 1/a + c + 1/a + b] - 3
S = 2016. 1/2016 - 3
S = - 2
Từ \(a+b+c=2016\) và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2016}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=2016.\frac{1}{2016}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=1\)
\(\Rightarrow\frac{\left(a+b\right)+c}{a+b}+\frac{\left(a+c\right)+b}{a+c}+\frac{\left(b+c\right)+a}{b+c}=1\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=1\)
\(\Rightarrow\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=-2\)
hay \(P=-2\)
