Prove that $\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3$ 
Prove that $a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3$
SOURCE : "A Brief Introduction to Olympiad Inequalities" by Evan Chen

It was one of the practice problems. Equality case is easy. I tried AM-GM and Muirhead, but could not seem to find a suitable proof for the inequality.
Any hint would be really helpful.
Thanks in advance !
^__^

Using $\text{AM-GM}$ this can be solved. Note that by $\text{AM-GM}$  $$4a^7+3b^7=a^7+a^7+a^7+a^7+b^7+b^7+b^7 \ge 7a^4b^3$$$$4b^7+3c^7=b^7+b^7+b^7+b^7+c^7+c^7+c^7 \ge 7b^4c^3$$$$4c^7+3a^7=c^7+c^7+c^7+c^7+a^7+a^7+a^7 \ge 7c^4a^3$$
Adding the three and dividing by seven, we have $$a^7+b^7+c^7 \ge a^4b^3+b^4c^3+c^4a^3$$
