Question

   Cho a^2+b^2+c^2=a.b+b.c+c.a

  CMR: a=b=c

Answer 1

gt <=>     \(a^2+b^2+c^2-ab-bc-ca=0\)

<=>     \(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

<=>    \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)        (1)

TA LUÔN CÓ:     \(\left(a-b\right)^2;\left(b-c\right)^2;\left(c-a\right)^2\ge0\forall a;b;c\)

=>     \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)        (2)

TỪ (1) VÀ (2) =>    DẤU "=" SẼ XẢY RA <=>     \(\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\)

<=>     \(a=b=c\)

VẬY TA CÓ ĐPCM.

Answer 2

a² + b² + c² = ab + bc + ca

<=> 2( a² + b² + c² ) = 2( ab + bc + ca )

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

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

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

Ta có : \(\hept{\begin{cases}\left(a-b\right)^2\\\left(b-c\right)^2\\\left(c-a\right)^2\end{cases}}\ge0\forall a,b,c\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Đẳng thức xảy ra ( tức là (*) xảy ra ) <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)

=> ĐPCM