Ta có : \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\Rightarrow\frac{a+b}{c}-1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có :
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)(vì a + b + c \(\ne\)0)
=> \(\hept{\begin{cases}\frac{a+b}{c}=2\\\frac{b+c}{a}=2\\\frac{c+a}{b}=2\end{cases}}\)=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)
Khi đó, ta có : \(\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right).\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)
Hay \(\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=\frac{2c.2b.2a}{a.c.b}=2.2.2=8\)
