Proving the inequality using am gm theorem

Question

I was asked to prove that

$a^3+b^3 \le (a^2+b^2)(a^4+b^4)$

I expanded the rhs and used am gm and got $a^6+a^2b^4+a^4b^2+b^6 \ge 4a^3b^3$ and struck.i think I lack intuition or I don't have enough experience.Any hints

Answer 1

Maybe you want to prove: $(a^3+b^3)^2 \le (a^2+b^2)(a^4+b^4)$. In this case, it reduces to: $2a^3b^3 \le a^2b^4+a^4b^2$, but this is true because $a^2b^4+a^4b^2 - 2a^3b^3 = (ab^2-a^2b)^2 \ge 0$ clearly true.