a2+b2+c2=ab+bc+ac
đề là thế này chí
Ta có : a2+b2+c2=ab+bc+ac
\(\Leftrightarrow\)2(a2+b2+c2)=2(ab+bc+ac)
\(\Leftrightarrow\)2a2+2b2+2c2-2ab-2bc-2ac=0
\(\Leftrightarrow\)a2+a2+b2+b2+c2+c2-2ab-2ac-2bc=0
\(\Leftrightarrow\)(a2-2ab+b2)+(a2-2ac+c2)+(b2-2bc+c2)=0
\(\Leftrightarrow\)(a-b)2+(a-c)2+(b-c)2=0
Ta có : \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0với\forall a,b\\\left(a-c\right)^2\ge0với\forall a,c\\\left(b-c\right)^2\ge0với\forall b,c\end{matrix}\right.\)
mà (a-b)2+(a-c)2+(b-c)2=0
\(\Leftrightarrow\)(a-b)2=(a-c)2=(b-c)2=0
\(\Leftrightarrow\)a-b=a-c=b-c=0
\(\Leftrightarrow\)a=b=c(đpcm)
