Ta có :
\(a^2+b^2+c^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2\left(ab+ba+ca\right)=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\)\(a^2+b^2+c^2=3\left(ab+bc+ca\right)-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\)\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow\)\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}}\)
Suy ra \(a=b=c\) ( đpcm )
Vậy \(a=b=c\)
Chúc bạn học tốt ~
\(a^2+b^2+c^2=3.\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2+2.\left(ab+ba+ca\right)=3.\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}}\)
\(\Rightarrow a=b=c\)
