Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it: 
https://imgur.com/user/Khansis/favorites/folder/7408635/math
 A: Hint:
$$
LHS-RHS=(b-a)^2(a^2+b^2)(3a^2+5ab+3b^2)
$$
A: It is $$4(a^6+b^6)-(a+b)(a^2+b^2)(a^3-b^3)=(a-b)^2(a^2+b^2)(3a^2+5ab+3b^2)\geq 0$$
A: Since the left side does not depend on changing signs of our variables, it's enough to prove our inequality for non-negatives $a$ and $b$.
Now, by C-S $$4(a^6+b^6)=2\cdot(1^2+1^2)(a^6+b^6)\geq2(a^3+b^3)^2.$$
Thus, it's enough to prove that:
$$2(a^3+b^3)\geq(a+b)(a^2+b^2)$$ or
$$2(a^2-ab+b^2)\geq a^2+b^2$$ or
$$(a-b)^2\geq0$$ and we are done!
A: You can simplify first:
$$4(\underbrace{a^6+b^6}_{(1)}) =4\require{cancel}\cancel{(a^2+b^2)}(a^4-a^2b^2+b^4)\ge (a+b)\cancel{(a^2+b^2)}(\underbrace{a^3+b^3}_{(2)}) \Rightarrow\\
4((a^2-b^2)^2+a^2b^2)\ge (a+b)^2((a-b)^2+ab) \Rightarrow \\
3(a^2-b^2)^2+ab(4ab-(a+b)^2)\ge 0 \Rightarrow \\
3(a-b)^2(a+b)^2-ab(a-b)^2\ge 0 \Rightarrow \\
(a-b)^2(\underbrace{3a^2+3b^2+5ab}_{(3)})\ge 0 \quad \color{green}\checkmark$$
The equality holds when $a=b$.
Note: 
1) For $(1),(2)$ it was used: $a^3+b^3=(a+b)(a^2-ab+b^2)$. 
2) For $(3)$, it was used AM-GM: $3(a^2+b^2)\ge 6|ab|\ge 5ab$.   
