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Thế này nhé ^^
- Ta có : \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\left(a+b+c\right)\left[\left(a^2+2ab+b^2\right)-bc-ac+c^2-3ab\right]\)
\(=\left[\left(a+b\right)+c\right].\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=a^3+b^3+c^3-3abc\)
- \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\frac{\left(a+b+c\right)}{2}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\right]=0\)
\(\Leftrightarrow\frac{\left(a+b+c\right)}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^2+b^2+c^2=ab+bc+ac\)
\(\Rightarrow2.\left(a^2+b^2+c^2\right)=2.\left(ab+bc+ac\right)\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) (BĐT luôn đúng)
Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\)
\(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(b^2-2ab+a^2\right)+\left(c^2-2bc+b^2\right)=0\)
\(\Leftrightarrow\left(a-c\right)^2+\left(b-a\right)^2+\left(c-b\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-c=0\\b-a=0\end{cases}}\) \(\Leftrightarrow a=b=c\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)(1)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}}\forall a,b,c\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\left(\forall a,b,c\right)\)(2)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Rightarrow a=b=c\)
Vậy \(a=b=c\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a=b=c\)
Có \(a^2+b^2=ab+ba\)
\(\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2-2ab+b^2=0\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a-b=0\Leftrightarrow a=b\)
\(\RightarrowĐPCM\)