Prove the inequality is true Here is a question that I need to prove 

Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$

So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$

If $a \ge b$, then $a^3 \ge b^3$ and $a^5 \ge b^5$, which implies $a^3 - b^3 \ge 0$ and $a^5 - b^5 \ge 0$. Since a positive times a positive is positive (or $0$ times anything is $0$), $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$
If instead $a < b$, then 
$a^3 < b^3$ and $a^5 < b^5$, which implies $a^3 - b^3 < 0$ and $a^5 - b^5 < 0$. Since a negative times a negative is positive, $$(a^3 - b^3)(a^5 -b^5) > 0.$$
Therefore, for all $a,b \ge 0$, $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$
