\(\dfrac{a^3}{b}+ab+\dfrac{b^3}{c}+bc+\dfrac{c^3}{a}+ca\ge2\sqrt{\dfrac{a^4b}{b}}+2\sqrt{\dfrac{b^4c}{c}}+2\sqrt{\dfrac{c^4a}{a}}=2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

áp dụng AM GM ta có a^3/b+ab>=2a^2

chứng minh tương tự => a^3/b+b^3/c+c^3/a>=2(a^2+b^2+c^2)-(ab+bc+ca)

mà ta có a^2+b^2+c^2>=(ab+bc+ca)

=>a^3/b+b^3/c+c^3/a>= ab+bc+ca

"=" xảy ra khi a=b=c

Được bạn nhé :"))))

Ủng hộ mình = cách theo dõi mình nha

a+b+c=0

\(\left(a+b+c\right)^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc=0\)

\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)+\left(3bc^2+3b^2c+3abc\right)-3abc=0\)\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+b+c\right)+3bc\left(a+b+c\right)-3abc=0\)\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

mk ko chắc cách bn đúng nhưng cách của mk là phù hợp nhất đó

Không nên chứng minh như thế này nhé. Ở ngay phần \(a+b=\frac{3abc}{-3ab}\) đã sai sót vì bạn không tính đến trường hợp \(a=0\) hoặc $b=0$ đã thực hiện phép chia như vậy.

Sử dụng hằng đẳng thức: \((a+b)^3=a^3+b^3+3ab(a+b)\) ta có:

\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3\)

Vì \(a+b+c=0\Rightarrow a+b=-c\). Thay vào biểu thức trên:

\((a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=-c^3+3abc+c^3=3abc\)

Do đó:

\(a^3+b^3+c^3=3abc\)

`(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`

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

`VT>=0`

Dấu "=" xảy ra khi `a=b=c`

`a^3+b^3+c^3=3abc`

`<=>a^3+b^3+c^3-3abc=0`

`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`

`<=>(a+b)^3+c^3-3ab(a+b+c)=0`

`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`

`**a+b+c=0`

`**a^2+b^2+c^2=ab+bc+ca`

`<=>a=b=c`

Ta có : \(a+b+c+d=0\)

\(\Leftrightarrow a+b=-c-d\)

\(\Leftrightarrow\left(a+b\right)^3=\left(-c-d\right)^3\)

\(\Leftrightarrow a^3+b^3+3ab.\left(a+b\right)=-c^3-d^3+3cd.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3cd.\left(c+d\right)-3ab.\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.cd.\left(a+b\right)+3ab.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.\left(c+d\right)\left(cd+ab\right)\)

Ta có : 

⇔a3+b3+c3+d3=3.cd.(a+b)+3ab.(c+d)

\(a^3+b^3+c^3-3abc\) \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ca-bc-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Vậy để \(a^3+b^3+c^3=3abc\) thì \(a+b+c=0\) hoặc \(a=b=c\)

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

b,

Ta có:

\(\left(a+b+c\right)^3=0\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)

a/ \(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^3=0\)

\(\Leftrightarrow\left[\left(a+b\right)+c\right]^3=0\)

\(\Leftrightarrow\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3bc^2+3b^2c+3a^2c+3ac^2+6abc=0\)

\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3bc^2+3b^2c+3abc\right)+\left(3ac^2+3a^2c+3abc\right)-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3+3abc\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)-3abc=0\)

Mà \(a+b+c=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)

b/ \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)

+) Nếu : \(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

Vậy \(a^3+b^3+c^3=3abc\) \(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

1. \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)

\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

2. \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

3.Còn có a + b + c = 0 nữa mà bn.

\(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)

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

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

1) Có: \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3-3abc=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

2)Có: \(a+b-c=0\)

\(\Leftrightarrow a+b=c\)

\(\Leftrightarrow\left(a+b\right)^3=c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3\)

\(\Leftrightarrow a^3+b^3+3abc=c^3\)

\(\Leftrightarrow a^3+b^3-c^3=-3abc\)

ta có : 

\(a^3+c^3=\left(a+c\right)^3-3ac\left(a+c\right)\)

nên \(a^3+c^3-b^3+3abc=\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)\)

\(=\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2-3ac\right]=\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)\)

b. tương tự ta có :

\(a^3-b^3-c^3-3abc=a^3-\left(b+c\right)^3+3bc\left(b+c-a\right)\)

\(=\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2-3bc\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)

c. ta có : \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=\left(x-z+z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left(x-z\right)^3+3\left(x-z\right)\left(z-y\right)\left(x-y\right)+\left(z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=3\left(x-z\right)\left(z-y\right)\left(x-y\right)\)