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
$$(a^3-b^3)(a^5-b^5)\geq 0$$
 A: Note that $$(a^3-b^3)(a^5-b^5) = (a-b)(a^2+ab+b^2)(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$
$$=(a-b)^2(a^2+ab+b^2)(a^4+a^3b+a^2b^2+ab^3+b^4)\ge 0$$
A: I don't think an expansion is needed at all. Here is an even more trivial proof.
Consider that if $a>b$,  $a^3 > b^3$ and $a^5> b^5$ since $a$ and $b$ are positive. 
So, we have that $$(a^3-b^3)(a^5-b^5) > 0$$ when $a>b\geq 0 \tag{1}$.
Now, if $a\leq b$, 
$b^5\geq a^5$ and $b^3 \geq a^3$
so we have that $$(b^3-a^3)(b^5-a^5) = (a^3-b^3)(a^5-b^5) \geq 0 \tag{2}$$ when $b\geq a\geq 0$. 
$(1)$ and $(2)$ are together necessary and sufficient condition to prove the required inequality, i.e. that $$\boxed{a^8+b^8 \geq a^3b^5 + a^5b^3} 
~~~\blacksquare$$
A: If $a\geq b$ then $a^3\geq b^3$ and $a^5\geq b^5$ so $(a^3-b^3)(a^5-b^5)\geq 0$. Otherwise $a^3-b^3$ is negative, and so is $a^5-b^5$, so their product is positive, and the inequality is proved.
A: 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 $a,b \ge 0$, $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$
A: WLOG, consider $a\ge b$. So:
$$a^3\ge b^3;a^5\ge b^5$$
Use Rearrangement inequality:
$$a^8+b^8\ge a^3b^5+b^3a^5.$$
