a: Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a+b+c=0\)
a) \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)(đúng do a+b+c = 0)
b) Ta có: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge ab+ac+bc\)
\(ĐTXR\Leftrightarrow a=b=c\), mà a,b,c đôi một khác nhau => Đẳng thức không xảy ra\(\Rightarrow a^2+b^2+c^2>ab+ac+bc\Rightarrow a^2+b^2+c^2-ab-ac-bc>0\)
Ta có: \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)\(\Rightarrow a+b+c=0\)( do (1))