How prove this inequality $a^4+b^4+c^4+6(a^2b^2+b^2c^2+a^2c^2)+4abc(a+b+c)<4(ab+bc+ac)(a^2+b^2+c^2)$ let $a,b,c>0$, and such
$$a^2+b^2+c^2<2ab+2bc+2ca$$
show that
$$a^4+b^4+c^4+6(a^2b^2+b^2c^2+a^2c^2)+4abc(a+b+c)<4(ab+bc+ac)(a^2+b^2+c^2)$$
I know this indentity:
$$a^2+b^2+c^2-2(ab+bc+ac)
=-(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})$$

Let $p = a+b+c, q = ab+bc + ca, r = abc$ then the inequality we need to show is equivalent to: 
$$p^4+16q^2 < 8p^2q+4pr$$ 
Using the condition, we know $p^2 < 4q$ and it is well known that $3q \le p^2$. Thus we have $(4q-p^2)(p^2-3q) \ge 0 \implies 7p^2q \ge p^4+12q^2$.
Using this, it is enough to show that $4q^2 < p^2q+4pr$.  But we always have $p^2q +3pr \ge 4q^2$, so this is true.

To show that $p^2q + 3pr \ge 4q^2$, i.e.
$$(a+b+c)^2(ab+bc+ca)+3(a+b+c)abc \ge 4(ab+bc+ca)^2$$
$$\iff \sum_{sym}a^3b \ge \sum_{sym} a^2b^2$$
which is evident using Muirhead or AM-GM as $a^3b+ab^3 \ge 2a^2b^2$.
