Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.

$$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6$$
But we know that $$(X-Y)^2\ge 0\Longleftrightarrow X^2+Y^2 \ge 2XY$$
taking $X= a^3$ and  $Y=b^3$ we get 
$$2a^3b^3 \le a^2b^4+b^2a^4 $$
so 
$$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6 \le a^6 + \color{red}{a^2b^4+b^2a^4 } + b^6  = (a^2+b^2)(a^4+b^4) $$
