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Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=2\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
\(\Leftrightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2=0\)
Vì \(\left\{{}\begin{matrix} -\left(a-b\right)^2\le0\\-\left(b-c\right)^2\le0\\-\left(c-a\right)^2\le0\end{matrix}\right.\Rightarrow-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\le0\)
Dấu ''= '' xảy ra \(\Leftrightarrow a=b=c\)
Vậy với a=b=c thì \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
a)a2+b2+c2+3=2(a+b+c)
=>a2+b2+c2+1+1+1-2a-2b-2c=0
=>(a2-2a+1)+(b2-2b+1)+(c2-2c+1)=0
=>(a-1)2+(b-1)2+(c-1)2=0
=>a-1=b-1=c-1=0 <=>a=b=c=1
-->Đpcm
b)(a+b+c)2=3(ab+ac+bc)
=>a2+b2+c2+2ab+2ac+2bc -3ab-3ac-3bc=0
=>a2+b2+c2-ab-ac-bc=0
=>2a2+2b2+2c2-2ab-2ac-2bc=0
=>(a2- 2ab+b2)+(b2-2bc+c2) + (c2-2ca+a2) = 0
=>(a-b)2+(b-c)2+(c-a)2=0
Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0
=>a-b hoặc b=c hoặc a=c
=>a=b=c
-->Đpcm
c)a2+b2+c2=ab+bc+ca
=>2(a2+b2+c2)=2(ab+bc+ca)
=>2a2+2b2+c2=2ab+2bc+2ca
=>2a2+2b2+c2-2ab-2bc-2ca=0
=>a2+a2+b2+b2+c2+c2-2ab-2bc-2ca=0
=>(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ca+c2)=0
=>(a-b)2+(b-c)2+(a-c)2=0
Hay (a-b)2=0 hoặc (b-c)2=0 hoặc (a-c)2=0
=>a-b hoặc b=c hoặc a=c
=>a=b=c
-->Đpcm
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)
b, Ta có \(m=a+b+c\)
\(\Rightarrow am+bc=a\left(a+b+c\right)+bc=a\left(a+b\right)+ac+bc=\left(a+c\right)\left(a+b\right)\)
CMTT \(bm+ac=\left(b+c\right)\left(b+a\right)\);\(cm+ab=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
(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-2ac-2bc=4a2+4b2+4c2-4ab-4ac-4bc
=>2a2+2b2+2c2-2ab-2ac-2bc-4a2-4b2-4c2+4ab+4bc+4ac=0
=>-2a2-2b2-2c2+2ab+2ac+2bc=0
=>-(2a2+2b2+2c2-2ab-2ac-2bc)=0
=>-[(a2-2ab+b2)+(b2-2bc+c2)+(a2-2ac+c2)]=0
=>-[(a-b)2+(b-c)2+(a-c)2]=0
=>(a-b)2+(b-c)2+(a-c)2=0
=>(a-b)=(b-c)=(a-c)=0
=>a-b=0 =>a=b (1)
b-c=0 =>b=c (2)
từ (1) và (2)
=>a=b=c (đpcm)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
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= \(2\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\)
=\(2\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]\)
=\(2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
=>\(\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]-2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow-1\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
=> a^2—2ab+b^2 +b^2-2bc+c^2+c^2-2ca+a^2-4a^2-4b^2-4c^2+4ab+4bc+4ca=0
=> —(2a^2+2^2+2c^2-2ab-2bc-2ca)=0
=>(a-b)^2+(b-c)^2+(c-a)^2=0
=>a=b;b=c;c=a
=>a=b=c