Xét (a+b+c)2 = 36
<=> a2 + b2 + c2 + 2 (ab+bc+ca) = 36
<=> ab + bc + ca = 12
Có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
<=> \(2a^2+2b^2+2c^2\ge2\left(ab+bc+ca\right)\)
<=> a2 + b2 + c2 \(\ge ab+bc+ca\)
Dấu "=" <=> a = b = c = 2
Mà \(a^2+b^2+c^2=ab+bc+ca=12\)
=> a = b = c = 2
\(12=6.4-12=4\left(a+b+c\right)-12\)
\(\Rightarrow a^2+b^2+c^2=4\left(a+b+c\right)-12\)
\(\Leftrightarrow\left(a^2-4a+4\right)+\left(b^2-4b+4\right)+\left(c^2-4c+4\right)=0\)
\(\Leftrightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c=2\)
