Làm theo kiểm phá giai đoạn nha.
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2+4abc>a^3+b^3+c^3\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2+4abc-a^3-b^3-c^3>0\)
\(\Leftrightarrow\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)>0\) đúng vì
\(\left\{{}\begin{matrix}a+b-c>0\\b+c-a>0\\c+a-b>0\end{matrix}\right.\)
\(\RightarrowĐPCM\)
Vì a, b, c là 3 cạnh tam giác nên:
\(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)>0\)
\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+c^2a+c^2b-2abc-a^3-b^3-c^3>0\)
\(\Leftrightarrow\left(ab^2-2abc+ac^2\right)+\left(ba^2-2abc+bc^2\right)+\left(ca^2-2abc+cb^2\right)+4abc>a^3+b^3+c^3\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2+4abc>a^3+b^3+c^3\) (ĐPCM)