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.

For example   $$4(a^2+b^2+c^2)^3-(a^3+b^3+c^3+3abc)^2=$$
$$=\sum_{cyc}(a^6+12a^4b^2+12a^4c^2-6a^3b^3-18a^4bc-a^2b^2c^2)\geq0,$$  which is  true by Muirhead.
