Prove that
$$3\sum_{sym}x^2y^2z+\sum_{sym}x^4y\ge 4\sum_{sym}x^3yz,\quad\forall x,y,z>0,$$
where $\sum_{sym}$ is the symmetric sum notation.
Context: I was reading about the Muirhead inequality and I was wondering if I take a convex combination between 2 symmetric sums $S_0$ and $S_1$: $S_{\alpha}=\alpha S_1+(1-\alpha)S_0$, where $S1\ge S\ge S_0$ and $S$ is also a symmetric sum, then is there a tactic to prove that $S_{\alpha}\le S$ or $S_{\alpha}\ge S$? I took the particular case in the problem and I tested it in a program, by generating many random triples, and I didn't find a counterexample. I tried to use AM-GM or to rewrite the inequality to use the Schur inequality, but without an improvement. I believe this type of inequalities are of interest for those who participate in math contests.
Best Answer
Because $$\sum_{sym}(3x^2y^2z+x^4y-4x^3yz)=\sum_{cyc}(x^4z+y^4z-x^3yz-y^3xz-6x^3yz+6x^2y^2z)=$$ $$=\sum_{cyc}(x-y)^2(z(x^2+xy+y^2)-3xyz)=\sum_{cyc}(x-y)^4z\geq0.$$