Chebyshev's sum inequality and a^a b^b c^c

1. For all $i,j$, the numbers $x_i-x_j$ and $y_i-y_j$ have the same sign, so $(x_i-x_j)(y_i-y_j)\ge0$. Summing over all ordered pairs $(i,j)$:
$$0\le\sum_{i,j}(x_i-x_j)(y_i-y_j)=2\cdot3\sum_ix_iy_i-2\left(\sum x_i\right)\left(\sum y_j\right).$$
Dividing by $2$ gives the claim.

2. Since $\ln$ is increasing, $(a,b,c)$ and $(\ln a,\ln b,\ln c)$ are similarly ordered. By part 1,
$$3(a\ln a+b\ln b+c\ln c)\ge(a+b+c)(\ln a+\ln b+\ln c).$$
Divide by $3$ and apply $\exp$:
$$a^ab^bc^c\ge(abc)^{(a+b+c)/3}.$$

Equality iff $a=b=c$.

Remark. The same proof works for any $n$ and any increasing function in place of $\ln$.

Lesson. When a sum looks like an "average of a product versus product of averages", Chebyshev is the tool.