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:
$$a^2+b^2+c^2-2(ab+bc+ac)
=-(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})$$
 A: $$
\because a,b,c > 0\\
(a+b+c)^4 = a^4 + b^4 + c^4 + 4(a^3 b + a^3 c + b^3 a + b^3 c + c^3 a + c^3 b) + 6(a^2 b^2 + a^2 c^2 + b^2 c^2) + 12abc(a+b+c) > a^4 + b^4 + c^4 + 6(a^2 b^2 + b^2 c^2 + a^2 c^2 ) + 4abc(a+b+c)$$
From your starting condition we get:
$$
a^2 + b^2 + c^2 - 2(ab+ac+bc) <0 \\
a^2 + b^2 + c^2 < 2(ab+ac+bc) \\
(a+b+c)^2 < 4(ab+ac+bc) \\
\therefore (a+b+c)^4 < 16(ab+ac+bc)^2 \\
\because 2(ab+ac+bc) < a^2 + b^2 + c^2 \\
\therefore (a+b+c)^4 < 16(ab+ac+bc)^2 < 4(a^2+b^2+c^2)^2
$$
Now just show that $16(ab+ac+bc)^2 < 4(ab+ac+bc)(a^2+b^2+c^2) < 4(a^2 + b^2 + c^2)^2$
A: 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 $a^3b+ab^3 \ge 2a^2b^2$.
