\(a^2+b^2+c^2\ge ab+bc+ca\)
<=>\(2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) luôn đúng => đpcm
Ta có: \(A^2+B^2+C^2\ge AB+AC+BC\)
\(\Leftrightarrow2\left(A^2+B^2+C^2\right)\ge2\left(AB+BC+AC\right)\)
\(\Leftrightarrow2A^2+2B^2+2C^2-2AB-2AC-2BC\ge0\)
\(\Leftrightarrow(A^2-2AB+B^2)+(A^2-2AC+C^2)+(B^2-2BC+C^2)\ge0\)
\(\Leftrightarrow\left(A-B\right)^2+\left(A-C\right)^2+\left(B-C\right)^2\ge0\) : luôn đúng vì
\(\left\{{}\begin{matrix}\left(A-B\right)^2\ge0\\\left(B-C\right)^2\ge0\\\left(A-C\right)^2\ge0\end{matrix}\right.\)
Vậy với mọi A,B,C ta đều có \(A^2+B^2+C^2\ge AB+AC+BC\).
