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: To be honest, I do not look at the answers before mine and tried it myself with the most innovative thought I am able to have:  $(a^3+b^3)^2 = (a^2\cdot a + b^2\cdot b)^2 \le (a^4+b^4)(a^2+b^2)$ by Cauchy-Schwarz inequality. As you can see how "beautiful" the CS inequality is....
A: $$\begin{array}{rrcl}
& (a^3+b^3)^2 &\le& (a^2+b^2)(a^4+b^4) \\
\iff& a^6 + 2a^3b^3 + b^6 &\le& a^6+a^2b^4+b^2a^4+b^6 \\
\iff& 2a^3b^3 &\le& a^2b^4+b^2a^4 \\
\iff& 2ab &\le& b^2+a^2 \\
\iff& 0 &\le& b^2-2ab+a^2 \\
\iff& 0 &\le& (b-a)^2 \\
\end{array}$$
A: $$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6$$
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 
$$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6 \le a^6 + \color{red}{a^2b^4+b^2a^4 } + b^6  = (a^2+b^2)(a^4+b^4) $$
A: \begin{eqnarray*}
\color{blue}{a^2b^2(a-b)^2} \geq 0 \\
\color{red}{a^6} +\color{blue}{a^4b^2+a^2 b^4} +\color{red}{b^6} \geq \color{red}{a^6} +\color{blue}{2 a^3b^3} +\color{red}{b^6} \\
(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2.
\end{eqnarray*}
