Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)\(\dfrac{a+b-c}{c}=1\Leftrightarrow\dfrac{a+b}{c}-\dfrac{c}{c}=1\Leftrightarrow\dfrac{a+b}{c}-1=1\Leftrightarrow\dfrac{a+b}{c}=2\)\(\dfrac{b+c-a}{a}=1\Leftrightarrow\dfrac{b+c}{a}-\dfrac{a}{a}=1\Leftrightarrow\dfrac{b+c}{a}-1=1\Leftrightarrow\dfrac{b+c}{a}=2\)\(\dfrac{c+a-b}{b}=1\Leftrightarrow\dfrac{c+a}{b}-\dfrac{b}{b}=1\Leftrightarrow\dfrac{c+a}{b}-1=1\Leftrightarrow\dfrac{c+a}{b}=2\)\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\\ =\dfrac{a+b}{a}\cdot\dfrac{b+c}{b}\cdot\dfrac{c+a}{c}\\ =\left(a+b\right)\cdot\dfrac{1}{a}\cdot\left(b+c\right)\cdot\dfrac{1}{b}\cdot\left(c+a\right)\cdot\dfrac{1}{c}\\ =\left(a+b\right)\cdot\dfrac{1}{c}\cdot\left(b+c\right)\cdot\dfrac{1}{a}\cdot\left(c+a\right)\cdot\dfrac{1}{b}\\ =\dfrac{a+b}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}\\ =2\cdot2\cdot2\\ =8\)
Vậy \(P=8\)
