# [Math] Prove the inequality is true

inequalityprovability

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 $$a,b \ge 0$$, $$(a^3 - b^3)(a^5 -b^5) \ge 0.$$