Question

Cho \(a^2+b^2+c^2-ab-ac-bc=0\). Chứng minh a=b=c

Answer 1

a² + b² + c² - ab - bc - ac = 0

⇔ a² - 2ab + b² + b² - 2bc + c² + c² - 2ac + a² = 0

⇔ ( a - b)² + ( b - c)² + ( c - a)² = 0

⇔ a - b = 0 ; b - c = 0 và c - a = 0

⇔ a = b ; b = c và c = a

⇔ a = b = c

Answer 2

Đặt: A= \(a^2+b^2+c^2-ab-ac-bc\)=0

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

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

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

Vì: \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

Dấu"=" xảy ra \(\Leftrightarrow a=b=c\)