\(P=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\\ =\dfrac{ab\left(a+b\right)}{abc}+\dfrac{bc\left(b+c\right)}{abc}+\dfrac{ca\left(c+a\right)}{abc}-\left(\dfrac{abc}{a}+\dfrac{abc}{b}\right)\left(\dfrac{abc}{b}+\dfrac{abc}{c}\right)\left(\dfrac{abc}{c}+\dfrac{abc}{a}\right)\\ =a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-\left(bc+ca\right)\left(ca+ab\right)\left(ab+bc\right)\\ =ab\left(a+b\right)+c\left(a^2+b^2\right)+c^2\left(a+b\right)-abc\left(a+b\right)\left(b+c\right)\left(c+a\right)\\ =ab\left(a+b\right)+c\left(a+b\right)^2-2abc+c^2\left(a+b\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)\\ =\left(a+b\right)\left[ab+c\left(a+b\right)+c^2\right]-\left(a+b\right)\left(b+c\right)\left(c+a\right)-2\\ =\left(a+b\right)\left(b+c\right)\left(c+a\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)-2\\ =-2\)
#$\mathtt{Toru}$
