Bài 2:
1/ \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3abc-3ab\left(a+b\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\) (đpcm).
Bài 3:
1/ \(M=a^3+b^3+ab=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=a^2-ab+b^2+ab=a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{1}{2}\)
Vậy \(MinM=\dfrac{1}{2}\)
