Muirhead's idea with weighted AM-GM

Idea. The exponent vector $(5,0,0)$ majorises $(3,1,1)$: each term $a^3bc$ is a weighted geometric mean of $a^5,b^5,c^5$.

Weighted AM-GM. With weights $\frac35,\frac15,\frac15$,
$$a^3bc=(a^5)^{3/5}(b^5)^{1/5}(c^5)^{1/5}\le\frac{3a^5+b^5+c^5}5.$$
The same holds for $b^3ca$ and $c^3ab$.

Summation.
$$abc(a^2+b^2+c^2)=a^3bc+b^3ca+c^3ab\le\frac{5(a^5+b^5+c^5)}5=a^5+b^5+c^5.$$
Equality iff $a=b=c$.

Remark. Muirhead's theorem says that $\sum_{\rm sym}a^{\alpha_1}b^{\alpha_2}c^{\alpha_3}\ge\sum_{\rm sym}a^{\beta_1}b^{\beta_2}c^{\beta_3}$ whenever $\alpha$ majorises $\beta$. Each instance can be proved exactly as above.

Lesson. An inequality between symmetric sums of monomials of the same degree is an application of weighted AM-GM: find the weights that turn the smaller monomial into a geometric mean of the larger ones.