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Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac-4a^2-4b^2-4c^2+4ab+4bc+4ac=0\)
\(\Leftrightarrow-2a^2-2b^2-2c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow-\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)(đpcm)
Có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2-4a^2-4b^2-4c^2+4ab+4ac+4bc=0\)
\(\Rightarrow-2a^2-2b^2-2c^2+2ab+2ac+2bc=0\)
\(\Rightarrow-\left(a^2-2ab+b^2\right)-\left(b^2-2bc+c^2\right)-\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(a-c\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\forall a;b;c}\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c}\)
Vậy \(a=b=c\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
<=>\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca\)\(=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
<=>\(0=2a^2+2b^2+2c^2-2ab-2bc-2ca\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\)<=> a-b=b-c=c-a <=> a=b=c
vế phải= \(2\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\)
=\(2\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]\)
=\(2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
=>\(\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]-2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow-1\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
(a - b)2 + (b - c)2 + (c - a)2 = 4(a2 + b2 + c2 - ab - ac - bc)
\(\Leftrightarrow\) a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ac + a2 = 4(a2 + b2 + c2 - ab - ac - bc)
\(\Leftrightarrow\) 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 4(a2 + b2 + c2 - ab - ac - bc)
\(\Leftrightarrow\) 2(a2 + b2 + c2 - ab - ac - bc) = 4(a2 + b2 + c2 - ab - ac - bc)
\(\Leftrightarrow\) a2 + b2 + c2 - ab - ac - bc = 2(a2 + b2 + c2 - ab - ac - bc)
\(\Leftrightarrow\) 0 = 2(a2 + b2 + c2 - ab - ac - bc) - (a2 + b2 + c2 - ab - ac - bc)
\(\Leftrightarrow\) a2 + b2 + c2 - ab - ac - bc = 0
\(\Leftrightarrow\) 2(a2 + b2 + c2 - ab - ac - bc) = 0
\(\Leftrightarrow\) 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
\(\Leftrightarrow\) (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ac + a2) = 0
\(\Leftrightarrow\) (a - b)2 + (b - c)2 + (c - a)2 = 0
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\a=c\\c=a\end{matrix}\right.\)
\(\Rightarrow\)a = b = c