giai,cho,minh,bai,nay,di cho,a,b,c>=0.CMR:a^3+b^3+c^3>=3abc
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Ta có :
Giả thuyết : a + b + c = 0
(a + b + c)3 = 0
a3 + b3 + c3 + 3.(a + b)(b + c)(c + a) = 0
Từ a + b + c = 0
=> \(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
=> a3 + b3 + c3 + 3.(-c)(-a)(-b) = 0
=> a3 + b3 + c3 = 3abc
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(a+b+c=0\Rightarrow-a-b=c\)
\(\Rightarrow a^3+b^3+c^3-3abc=a^3+b^3+\left(-a-b\right)^3-3abc\)
\(=a^3+b^3-a^3-3a^2b-3ab^2-b^3-3abc\)
\(\Rightarrow-3a^2b-3ab^2-3abc=3ab\left(-a-b\right)-3abc\)
\(=3abc-3abc=0\) (đpccm)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)
\(\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3=0\)
\(a^3+b^3+3ab\left(a+b\right)+c^3=0\)
\(a^3+b^3+c^3+3ab\left(-c\right)=0\)
\(a^3+b^3+c^3=3abc\)
Ta có:\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+3c^2-3ab\right)\)
\(=0\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(dpcm\right)\)
a+b+c=0
a+b=-c
(a+b)^3=(-c)^3
a^3+3a^2b+3ab^2+b^3=(-c)^3
a^3+b^3+c^3=-3a^2b-3ab^2
a^3+b^3+c^3=-3ab(-c)
a^3+b^3+c^3=3abc