nhân 2 với cả 2 vế
\(a^2+b^2+c^2\ge ab+bc+ac\left(1\right)\\ < =>2a^2+2b^2+2c^2\ge2ab+2bc+2ac\\ < =>\left(a^2-ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2+2ac+c^2\right)\ge0\\ < =>\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\left(2\right)\)
có
\(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{matrix}\right.\)
=> (2) luôn đúng
=> (1) luôn đúng (dấu '=' xảy ra khi a = b = c)
chúc may mắn :)
\(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2ac\ge0\) \(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\ge0\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Đúng với mọi a , b
Đẳng thức xảy ra khi \(\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Rightarrow a=b=c\)
theo Cô-si thì:
\(a^2+b^2\ge2ab\)
\(b^2+c^2\ge2bc\)
\(a^2+c^2\ge2ac\)
Cộng ba vế lại ta được:
\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
=>\(a^2+b^2+c^2\ge ab+bc+ac\) (đpcm)
