Biến đổi tương đương :
\(\left|a-b\right|+\left|b-c\right|+\left|c-a\right|\ge\sqrt{a^2+b^2+c^2-ab-bc-ac}\)
\(\Leftrightarrow4\left|a-b\right|+4\left|b-c\right|+4\left|c-a\right|\ge\sqrt{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(\Leftrightarrow4\left|a-b\right|+4\left|b-c\right|+4\left|c-a\right|\ge\sqrt{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
Đặt \(\left|a-b\right|=x;\left|b-c\right|=y;\left|c-a\right|=z\)
\(BĐT\Leftrightarrow4x+4y+4z\ge\sqrt{x^2+y^2+z^2}\)
\(\Leftrightarrow16\left(x^2+y^2+z^2+2xy+2yz+2xy\right)\ge x^2+y^2+z^2\)
\(\Leftrightarrow15x^2+15y^2+15z^2+32xy+32yz+32xz\ge0\) (luôn đúng vì \(x;y;z\ge0\))
Vậy \(\left|a-b\right|+\left|b-c\right|+\left|c-a\right|\ge\sqrt{a^2+b^2+c^2-ab-bc-ac}\)
