Đặt [tex]a-b=X; b-c=Y; c-a= Z\Rightarrow X+Y+Z=0[/tex]
Ta có:
[tex](\frac{1}{X}+\frac{1}{Y}+ \frac{1}{Z})^2= \frac{1}{X^2}+\frac{1}{Y^2}+\frac{1}{Z^2}+\frac{2}{XY}+\frac{2}{YZ}+\frac{2}{ZX}= \frac{1}{X^2}+\frac{1}{Y^2}+\frac{1}{Z^2}+ 2.\frac{X+Y+Z}{XYZ}= \frac{1}{X^2}+\frac{1}{Y^2}+\frac{1}{Z^2}[/tex]
Nên [tex]\sqrt{ \frac{1}{X^2}+\frac{1}{Y^2}+\frac{1}{Z^2}}=\sqrt{(\frac{1}{X}+\frac{1}{Y}+\frac{1}{Z})^2}=\left | \frac{1}{X}+\frac{1}{Y}+\frac{1}{Z} \right |[/tex]
Suy ra [tex]\sqrt{\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}}=\left | \frac{1}{a-b} +\frac{1}{b-c}+\frac{1}{c-a}\right |[/tex]
Vì a,b,c là số hữu tỉ nên[tex]\left | \frac{1}{a-b} +\frac{1}{b-c}+\frac{1}{c-a}\right |[/tex] cũng là số hữu tỉ
=> đpcm