Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$ 
Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$.

This question reminds me of rearrangement, but I can't really find two sequences that fit. Maybe there is a way using the triangle inequality, but I am unsure.

Also, we can use SOS:
$$\sum_{cyc}(a^5-a^3bc)=\sum_{cyc}a^3(a^2-bc)=$$
$$=\frac{1}{2}\sum_{cyc}a^3((a-b)(a+c)-(c-a)(a+b))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)(a^3(a+c)-b^3(b+c))=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)(a^2+b^2)+c(a^2+ab+b^2))\geq0.$$
