+)Ta có :\(\frac{a}{a^,}+\frac{b^,}{b}=1\) \(\iff\) \(ab+a^,b^,=a^,b\) \(\iff\) \(abc+a^,b^,c=a^,b^,c\left(1\right)\)
+)Ta có :\(\frac{b}{b^,}+\frac{c^,}{c}=1\)\(\iff\) \(bc+b^,c^,=b^,c\)\(\iff\) \(a^,bc+a^,b^,c^,=a^,b^,c^,\left(2\right)\)
Cộng \(\left(1\right)\) với \(\left(2\right)\) vế với vế ta được: \(abc+a^,b^,c+a^,bc+a^,b^,c^,=a^,bc+a^,b^,c\)
\(\implies\)\(abc+a^,b^,c^,=0\left(đpcm\right)\)