Proving the inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$ If $a,b,c$ are non-negative real numbers prove the following inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$.

Expanding, this is equivalent to (with $\sum$ denoting cyclic sums):
$$4\sum a^4  +3 abc\sum a \ge 2\sum ab(a^2+b^2)+3\sum a^2b^2 $$
which follows from combining Schur's inequality of fourth degree
$$3\sum a^4+3abc\sum a \ge 3\sum ab(a^2+b^2)$$
with the AM GMs:
$$\sum a^4 \ge \sum a^2b^2, \quad \sum ab(a^2+b^2) \ge 2\sum a^2b^2$$
