khai triển cái VT ra bạn:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Vì đề là a;b;c không âm,ta áp dụng bđt AM-GM:
\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\b+c\ge2\sqrt{bc}\\c+a\ge2\sqrt{ac}\end{matrix}\right.\) \(\Leftrightarrow\left(a+b+c\right)^3\ge a^3+b^3+c^3+3.2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=VP\)
\("="\Leftrightarrow a=b=c\)
Ta có:\(\left(a+b+c\right)^3=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
=\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Cần cm:\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Lại có:\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân vế theo vế ta có đpcm
Dấu "=" xảy ra khi a=b=c
