Question

cho (a-b)^2 + (b-c)^2 + (c-a)^2=4*(a^2+b^2+c^2-a*b-a*c-b*c) cmr:a=b=c

Answer 1

(a-b)²+(b-c)²+(c-a)²=4(a²+b²+c²-ab-ac-bc)

<=>a²-2ab+b²+b²-2bc+c²+c²-2ac+a²=4a²+4b²+4c²-4ab-4ac-4bc

<=>2a²+2b²+2c²-2ab-2bc-2ac-4a²-4b²-4c²+4ab+4ac+4bc=0

<=>2ab+2ac+2bc-2a²-2b²-2c²=0

<=>-[(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)]=0

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

<=>(a-b)²+(b-c)²+(c-a)²=0

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

Dấu "=" xảy ra <=> \(\left(a-b\right)^2=\left(b-c\right)^2=\left(a-c\right)^2=0\)

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

<=>a=b=c(đpcm)