Không mất tính tổng quát, giả sử \(a\le b\le c\)
Do \(a^2+b^2+c^2=1\Rightarrow a^2\le b^2\le c^2\le1\Rightarrow a\le b\le c\le1\)
\(\Rightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\\1-c\ge0\end{matrix}\right.\)
\(a^2+b^2+c^2=a^3+b^3+c^3\)
\(\Leftrightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)=0\)
Mà \(\left\{{}\begin{matrix}a^2\left(1-a\right)\ge0\\b^2\left(1-b\right)\ge0\\c^2\left(1-c\right)\ge0\end{matrix}\right.\) \(\Rightarrow\) dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}a^2\left(1-a\right)=0\\b^2\left(1-b\right)=0\\c^2\left(1-c\right)=0\\a^2+b^2+c^2=1\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(0;0;1\right)\)
\(\Rightarrow a^2+b^{2012}+c^{2013}=1\)
