Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2]$$
Since $\sum_{sym} x^3+3xyz-\sum_{sym} xy(x+y)\ge 0$, set $(x,y,z)\rightarrow (ab,bc,ca)$ we obtain $$\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\ge 0$$
Thus $LHS-RHS\ge 0$
I think this's a complicated solution and hard to find it by hand :">
Please give me a simplier solution. Thank you very much.

We have$:$ \begin{align*} \text{LHS}-\text{RHS}\\&=\frac{1}{3} \sum\limits_{cyc} \left( 2\,{a}^{4}+10\,{a}^{2}{b}^{2}+2\,{a}^{2}bc+5\,{a}^{2}{c}^
{2}+2\,a{b}^{3}+4\,ab{c}^{2}+{b}^{4}+4\,{b}^{3}c+24\,{c}^{4} \right) 
 \left( a-b \right) ^{2}\end{align*}
Another solution.
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ We have$:$
\begin{align*} \text{LHS}-\text{RHS}&=\frac{4}{3}\, \left( 5\,{p}^{2}-9\,q \right)  \left( {q}^{2} -3\,pr\right) \\&\,\,\,+\frac{1}{3}\, \left( {p}^{2}-3\,q \right)  \left( 3\,{p}^{4}-9\,{p}^{2}q+4\,{q} ^{2} \right)\\&\,\,\, -16\,{p}^{3}r+4\,{p}^{2}{q}^{2}+72\,pqr-16\,{q}^{3}-108\, {r}^{2}\\&\geqq 0\end{align*}
Note that$:$ $$-16\,{p}^{3}r+4\,{p}^{2}{q}^{2}+72\,pqr-16\,{q}^{3}-108\, {r}^{2}=4(a-b)^2(b-c)^2(c-a)^2 \geqq 0$$
