từ bài ra ta có : \(a\ne b;b\ne c;c\ne a\)
= \(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\ne0\)
=\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=0\)
=>\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a}{\left(b-c\right)\left(a-b\right)}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\)
\(\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)
=>\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a\left(c-a\right)+a\left(a-b\right)+b\left(b-c\right)+b\left(a-b\right)+c\left(c-a\right)+c\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
=>\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
=>\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
=> 3 số a,b,c không cùng âm và không cùng dương
=> trong 3 số a,b,c có ít nhất 1 số âm và 1 số dương
