Let $T_{m,n,p}(x,y,z)=\sum_{Sym} x^m y^n z^p$.
For $x,y,z>0$, prove
$2T_{6,3,0}(x,y,z)+T_{3,3,3}(x,y,z)+3T_{4,4,1}(x,y,z)\geq 6T_{5,2,2}(x,y,z)$.
I tried to prove that by using AM-GM inequality, without success.
Is there a general way to prove these "Muirhead-like" inequalities?
This is a very old question, but I'll answer it anyway.
We should prove that:
$$ \sum_{sym}2x^6 y^3 + x^3 y^3 z^3 + 3 x^4 y^4 z^1 - 6 x^5 y^2 z^2 \ge 0$$
If we divide the inequality by $x^3 y^3 z^3$, we get:
$$ \sum_{sym} 2 \frac{x^3}{z^3} + 1 + 3 \frac{x y}{z^2} - 6 \frac{x^2}{y z} \ge 0$$
Apply the following subsitution:
$$\frac{x}{z}=a, \frac{z}{y}=b, \frac{y}{x}=c$$
The inequality becomes:
$$ \sum_{sym} 2 a^3 + 1 + 3 \frac{a}{b} - 6 \frac{a}{c} \ge 0$$
Note that $abc=1$ so we can homogenize the last inequality in the following way:
$$ \sum_{sym} 2 a^3 + abc + 3 \frac{a}{b} abc - 6 \frac{a}{c} abc \ge 0$$
$$ \sum_{sym} 2 a^3 + abc + 3 a^2 c - 6 a^2b \ge 0$$
If we note that:
$$ \sum_{sym}a^2 b \equiv \sum_{sym}a^2 c $$
...the inequality reduces to:
$$ \sum_{sym} 2 a^3 + abc - 3 a^2b \ge 0$$
...or, by using "Muirhead-like" notation:
$$2T_{3, 0, 0} + T_{1, 1, 1} - 3T_{2, 1, 0} \ge 0$$
...which is obvious, because:
$$T_{3, 0, 0} \ge T_{2, 1, 0} \quad (Muirhead)$$
$$T_{3, 0, 0} + T_{1, 1, 1} \ge T_{2, 1, 0} \quad (Schur)$$
