Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$

Question

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

Answer 1

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!

Answer 2

Hint:

$$ LHS-RHS=(b-a)^2(a^2+b^2)(3a^2+5ab+3b^2) $$

Answer 3

It is $$4(a^6+b^6)-(a+b)(a^2+b^2)(a^3-b^3)=(a-b)^2(a^2+b^2)(3a^2+5ab+3b^2)\geq 0$$

Answer 4

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: $3(a^2+b^2)\ge 6|ab|\ge 5ab$.