Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ac-bc-ca\right)\)
⇔ \(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ac-4bc-4ca\)
⇔ \(2a^2+2b^2+2c^2-2ac-2bc-2ca=4a^2+4b^2+4c^2-4ac-4bc-4ca\)
⇔ \(2a^2+2b^2+2c^2-2ac-2bc-2ca=0\)
⇔ \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\) \(\forall a,b,c\)
⇒ \(\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
⇔ \(a=b=c\)
⇒ \(ĐPCM\)
