Để đỡ khó nhìn, ta đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\) \(\Rightarrow x+y+z=0\) và \(x;y;z\in Q\)
\(A=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}}\)
\(A=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{xz}+\frac{2}{yz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(A=\left|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right|\Rightarrow A\) hữu tỉ