Proving a cyclic inequality

Question

Show that $a^4 + b^4 + c^4 \geq a^3b + b^3c + c^3a$ for any postive integers $a, b, c$
I'm not sure how to approach this problem. I've tried assuming that WLOG $a > b > c$ so that it is clear that $$a^4 > a^3b,$$ $$b^4 > b^3c,$$ $$c^4 < c^3a $$ before realising that it does not get me anywhere, any ideas?

Answer 1

Also, we can use AM-GM here: $$\sum_{cyc}a^4=\frac{1}{4}\sum_{cyc}(3a^4+b^4)\geq\sum_{cyc}\sqrt[4]{(a^4)^3b^4}=\sum_{cyc}a^3b.$$

Answer 2

You can not assume $a\geq b\geq c$ because our inequality is not symmetric.

$(a^3,b^3,c^3)$ and $(a,b,c)$ have the same ordering.

Thus, by Rearrangement $$\sum_{cyc}a^4=\sum_{cyc}(a^3\cdot a)\geq\sum_{cyc}a^3b.$$

Answer 3

Also, we can use SOS: $$\sum_{cyc}(a^4-a^3b)=\sum_{cyc}\left(a^4-a^3b-\frac{1}{4}(a^4-b^4)\right)=$$ $$=\frac{1}{4}\sum_{cyc}(a-b)^2(3a^2+2ab+b^2)\geq0.$$