\(< =>\left(a+b\right)^2+2\left(a+b\right)c+c^2-3ab-3bc-3ca=0\)
\(< =>a^2+2ab+b^2+2ac+2bc-3ab-3bc-3ca=0\)
\(< =>a^2+b^2+c^2-ab-bc-ca=0\)
\(< =>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(< =>a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2=0\)
\(< =>\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
dấu"=" xảy ra<=>a=b=c
Ta có :
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)-3\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+ab+bc+ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ac=0\)(nhân 2 vế cho 2)
\(\Leftrightarrow\)\(a^2+b^2+c^2+\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2=0\)
\(\Rightarrow\)\(\left\{{}\begin{matrix}a^2=0,b^2=0,c^2=0\\\left(a+b\right)^2=0\\\left(b+c\right)^2=0\\\left(a+c\right)^2=0\end{matrix}\right.\Leftrightarrow a=b=c=0\left(đcpcm\right)\)
Ta có: \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc-3ab-3bc-3ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)