Ta có: a2+b2+c2=ab+bc+ca
=>2(a2+b2+c2)=2(ab+bc+ca)
<=>2a2+2b2+2c2=2ab+2bc+2ca
<=>2a2+2b2+2c2-2ab-2bc-2ca=0
<=>a2+a2+b2+b2+c2+c2-2ab-2bc=2ca=0
<=>(aa-2ab+b2)+(b2-2bc+b2)+(a2-2ca+c2)=0
<=>(a-b)2+(b-c)2+(a-c)2=0
=>hoặc (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0<=>a-b=0 hoặc b-c=0 hoặc a-c=0<=>a=b hoặc b=c hoặc a=c
=> a=b=c (đpcm)
a2+b2+c2=ab+bc+ac
<=>2(a2+b2+c2)=2(ab+bc+ac)
<=>2a2+2b2+2c2=2ab+2bc+2ac
<=> 2a2+2b2+2c2-(2ab+2bc+2ac)=0
<=> 2a2+2b2+2c2-2ab-2bc-2ac=0
<=> (a2-2ab+b2)+(b2-2bc+c2)+(a2-2ac+c2)=0
<=> (a-b)2+(b-c)2+(a-c)2=0
\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}\Leftrightarrow}a=b=c\left(dpcm\right)}\)
