Muirhead's inequality

Notation. For a symmetric sum, $T_{(\lambda_1,\lambda_2,\lambda_3)}$ is the sum over all distinct permutations of $a^{\lambda_1}b^{\lambda_2}c^{\lambda_3}$ divided by the number of permutations.

Majorization. $(3,0,0)\succ(2,1,0)$ because $3\ge2$, $3+0\ge2+1=3$ (equality), $3+0+0=2+1+0$ (equality). The inequality $3>2$ in the first component is the strict part.

Muirhead. $(3,0,0)\succ(2,1,0)$ implies $T_{(3,0,0)}\ge T_{(2,1,0)}$, i.e.
$$\frac{a^3+b^3+c^3}{3}\ge\frac{a^2b+a^2c+ab^2+b^2c+ac^2+bc^2}{6}.$$
Multiplying by $6$:
$$2(a^3+b^3+c^3)\ge a^2b+a^2c+ab^2+b^2c+ac^2+bc^2.$$

Conclusion. The claimed inequality holds, with equality iff $a=b=c$.