\(\Leftrightarrow a^3+b^3+abc-a^2b-ab^2-abc\ge0\)

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

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

a: Ta có: \(a+b+c=0\)

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

Ta có: a+b+c=0

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

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

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

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

b: Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

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

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

b) Ta có: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge ab+ac+bc\)

\(ĐTXR\Leftrightarrow a=b=c\), mà a,b,c đôi một khác nhau => Đẳng thức không xảy ra\(\Rightarrow a^2+b^2+c^2>ab+ac+bc\Rightarrow a^2+b^2+c^2-ab-ac-bc>0\)

Ta có: ​\(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)​\(\Rightarrow a+b+c=0\)( do (1))

a: Ta có: a+b+c=0

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

Ta có: a+b+c=0

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

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

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

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

b: Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

ai cứu mình với ạ:(

Ta có hằng đẳng thức: 

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=0.\left(a^2+b^2+c^2-ab-bc-ca\right)+3.1=0+3=3\)

Áp dụng bđt Cô- si với các số a,b,c>0:

\(a^3+1+1\ge3a,b^3+1+1\ge3b,c^3+1+1\ge3c\)

\(\Rightarrow a^3+b^3+c^3+6\ge3a+3b+3c\) 

\(\Rightarrow a^3+b^3+c^3\ge a+b+c+2\left(a+b+c\right)-6\ge a+b+c+2\cdot3\sqrt[3]{abc}-6=a+b+c+6-6=a+b+c\)

Vậy...

đề là chứng minh bất đẳng thức

bạn nào giải được giúp mình với, mình cảm ơn !

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

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

=>\(2\left(ab+bc+ac\right)=0\)

=>ab+bc+ac=0

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)

=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)

\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)

=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)

=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)

=>0=0(đúng)

Do a;b;c là độ dài 3 cạnh của 1 tam giác nên \(a;b;c>0\)

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

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

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

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

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

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

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

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

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

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

Hay tam giác ABC đều