\(8(a^3 + b^3+c^3)\ge (a+b)^3 + (b+c)^3 + (c+a)^3 \)

\(\Leftrightarrow 8(a^3 +b^3+c^3) \ge 2(a^3+b^3+c^3) + 3ab(a+b) + 3bc(b+c)+3ca(c+a) \)

\(\Leftrightarrow\)\(6(a^3 +b^3+c^3) \ge 3[ab(a+b) + bc(b+c) + ca(c+a)] \)

\(\Leftrightarrow\) \(2a^3+2b^3+2c^3 \ge ab(a+b) + bc(b+c) + ca(c+a) \)

We have: a³ + b³ \(\ge \) a²b + ab²

Prove that the same way as the way of Kaya Renger to prove that: \(a^4+b^4 \ge a^3b+ab^3\)

So \(\left\{{}\begin{matrix}a^3+b^3\ge ab\left(a+b\right)\\b^3+c^3\ge bc\left(b+c\right)\\c^3+a^3\ge ca\left(c+a\right)\end{matrix}\right.\)

So true. 

=> \(8(a^3 + b^3+c^3)\ge (a+b)^3 + (b+c)^3 + (c+a)^3 \)

8(a3+b3+c3)≥(a+b)3+(b+c)3+(c+a)3

⇔8(a3+b3+c3)≥2(a3+b3+c3)+3ab(a+b)+3bc(b+c)+3ca(c+a)

⇔

6(a3+b3+c3)≥3[ab(a+b)+bc(b+c)+ca(c+a)]

⇔

 2a3+2b3+2c3≥ab(a+b)+bc(b+c)+ca(c+a)

We have: a3 + b3 ≥

 a2b + ab2

Prove that the same way as the way of Kaya Renger to prove that: a4+b4≥a3b+ab3

So ⎧⎩⎨⎪⎪a3+b3≥ab(a+b)b3+c3≥bc(b+c)c3+a3≥ca(c+a)

So true. 

=> 8(a3+b3+c3)≥(a+b)3+(b+c)3+(c+a)3