Đáp án:

`a=b=c` 

Giải thích các bước giải:

`(a-b)^2+(b-c)^2+(c-a)^2=(a+b-2c)^2+(b+c-2a)^2+(c+a-2b)^2`
`<=>(a-b)^2+(b-c)^2+(c-a)^2-(a+b-2c)^2-(b+c-2a)^2-(c+a-2b)^2=0`
`<=>[(a-b)^2-(a+b-2c)^2]+[(b-c)^2-(b+c-2a)^2]+[(c-a)^2-(c+a-2b)^2]=0`
`<=>(a-b-a-b+2c)(a-b+a+b-2c)+(b-c-b-c+2a)(b-c+b+c-2a)+(c-a-c-a+2b)(c-a+c+a-2b)=0`
`<=>[(a-a)-(b+b)+2c][(a+a)+(b-b)-2c]+[(b-b)-(c+c)+2a][(b+b)+(c-c)-2a]+[(c-c)-(a+a)+2b][(c+c)+(a-a)-2b]=0`
`<=>(2c-2b)(2a-2c)+(2a-2c)(2b-2a)+(2b-2a)(2c-2b)=0`
`<=>2.(c-b).2.(a-c)+2.(a-c).2.(b-a)+2.(b-a).2.(c-b)=0`
`<=>4(c-b)(a-c)+4(a-c)(b-a)+4(b-a)(c-b)=0`
`<=>4.[(c-b)(a-c)+(a-c)(b-a)+(b-a)(c-b)]=0`
`<=>(c-b)(a-c)+(a-c)(b-a)+(b-a)(c-b)=0`
`<=>ac-c^2-ab+bc+ab-a^2-bc+ac+bc-b^2-ac+ab=0`
`<=>(ac+ac-ac)+(-ab+ab+ab)+(bc-bc+bc)-a^2-b^2-c^2=0`
`<=>-a^2-b^2-c^2+ac+ab+bc=0`
`<=>-(a^2+b^2+c^2-ac-ab-bc)=0`
`<=>a^2+b^2+c^2-ac-ab-bc=0`
`<=>2.(a^2+b^2+c^2-ac-ab-bc)=0`
`<=>2a^2+2b^2+2c^2-2ac-2ab-2bc=0`
`<=>(a^2-2ac+c^2)+(b^2-2ab+a^2)+(c^2-2bc+b^2)=0`
`<=>(a-c)^2+(b-a)^2+(c-b)^2=0`
Ta có:
`{((a-c)^2ge0foralla;c),((b-a)^2ge0forallb;a),((c-b)^2ge0forallx;b):}`
`=>(a-c)^2+(b-a)^2+(c-b)^2ge0foralla;b;c`
Mà `(a-c)^2+(b-a)^2+(c-b)^2=0`
`<=>{((a-c)^2=0),((b-a)^2=0),((c-b)^2=0):}`
`<=>{(a-c=0),(b-a=0),(c-b=0):}`
`<=>{(a=c),(b=a),(c=b):}`
`<=>a=b=c(text{ĐPCM})`
Vậy `a=b=c`

$\\$

`(a-b)^2 + (b-c)^2 + (c-a)^2 =(a+b-2c)^2 + (b+c-2a)^2 + (c+a-2b)^2`

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

`⇔ [(a-b)^2 - (a+b-2c)^2] + [(b-c)^2 - (b+c-2a)^2] + [(c-a)^2 - (c+a-2b)^2]=0`

`⇔ [(a-b-a-b+2c) (a-b+a+b-2c)] + [(b-c-b-c+2a) (b-c+b+c-2a)] + [(c-a-c-a+2b) (c-a+c+a-2b)]=0`

`⇔ [ (-2b+2c) (2a - 2c)] + [(-2c+2a) (2b-2a)] + [(-2a+2b) (2c-2b)]=0`

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

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

`⇔ 4 (c-b)(a-c) + 4 (a-c) (b-a) + 4 (b-a)(c-b)=0`

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

`⇔ ac - c^2 - ab + bc + ab - a^2 - bc + ac + bc - b^2 -ac + ab=0`

`⇔ -a^2 - b^2 -c^2 + ab + ac + bc=0`

`⇔ a^2 + b^2 +c^2 -ab-ac-bc=0`

`⇔ (a^2-2ab + b^2)+(a^2 - 2ac + c^2) + (b^2 - 2bc +c^2)=0`

`⇔ (a-b)^2 + (a-c)^2 + (b-c)^2=0`

Mà `(a-b)^2 + (a-c)^2 + (b-c)^2 ≥0∀a,b,c`

Dấu "`=`" xảy ra khi :

`(a-b)^2=0, (a-c)^2=0, (b-c)^2=0`

`⇔ a-b=0,a-c=0, b-c=0`

`⇔a=b,a=c,b=c`

`⇔a=b=c`