b, Ta có \(m=a+b+c\)

          \(\Rightarrow am+bc=a\left(a+b+c\right)+bc=a\left(a+b\right)+ac+bc=\left(a+c\right)\left(a+b\right)\)

CMTT \(bm+ac=\left(b+c\right)\left(b+a\right)\);\(cm+ab=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

a²+b²+c²=ab+bc+ac

<=>2a²+2b²+2c²=2ab+2bc+2ac

<=>2a²+2b²+2c²-2ab-2bc-2ac=0

<=>a²-2ab+b²+a²-2ac+c²+b²-2bc+c²=0

<=>(a-b)²+(a-c)²+(b-c)²=0

<=> a-b=b-c=a-c=0

<=>a=b=c

ko hiểu chỗ nào thì hỏi

\(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\) 

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

\(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(=\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(a-c\right)^2\ge0\\\left(b-c\right)^2\ge0\end{cases}}\Rightarrow\)\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)

Vậy a = b = c (đpcm)

a² + b² + c² = ab + bc + ca

<=> 2(a² + b² + c²) = 2(ab + bc + ca)

<=> a² - 2ab+ b² + c² - 2ca+ a² + b² - 2bc+ c² = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> a = b = c

a) => 2a^2 + 2b^2 = 2ab + 2ba

=>  2a^2 + 2b^2 - 2ab - 2ba = 0

=> (a-b)^2 + (a-b)^2 = 0

=> 2(a-b)^2 = 0

=> a-b = 0

=> a = b

b) Nhân hai vế với 2 và làm tương tự câu a)

=> (a-b)^2 + (b-c)^2 + (a-c)^2 = 0

=> a = b = c