Question

Answer 1

\(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)^2\)
\(3a^2+3b^2+3c^2=a^2+b^2+c^2+2ab+2ac+2bc\)
\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
mà \(\left(a-b\right)^2,\left(a-c\right)^2,\left(b-c\right)^2\ge0\forall a,b,c\)
\(\Rightarrow\begin{cases}\left(a-b\right)^2=0\\ \left(a-c\right)^2=0\\ \left(b-c\right)^2=0\end{cases}\)
\(\begin{cases}a-b=0\\ a-c=0\\ b-c=0\end{cases}\)
\(\begin{cases}a=b\\ a=c\\ b=c\end{cases}\)
\(a=b=c\left(dpcm\right)\)