There must be some condition missing. If we put for exemple $(a,b,c)=(-1,-1,1)$, the ineq is equivalent to $(\frac{1}{n}\sum_{i=1}^nx_i^3)\geq (\frac{1}{n} \sum_{i=1}^nx_i^2)(\frac{1}{n}\sum_{i=1}^nx_i)$ which is true by Chebyshev. But if we add the condition $c\le 0$, then it can be easily shown: Denote the $j$-th power mean by $M_j$, so the ineq reads $M_1^aM_2^{2b}M_3^{3c}\ge1$. If this is to be valid for all $x_i$, it must be homogenous so $a+2b+3c=0.$ So all feasible vectors are part of a 2-dimensional space. The vectors $(-2,1,0)$ and $(-1,2,-1)$ are feasible, as for the first one the ineq is equivalent to $M_2^2\ge M_1^2$ which is obvious, and for the second one it is equivalent to $M_2^2\ge M_1M_3$ which follows by the well-known concavity of the power means. If two vectors are feasible, so is any positive linear combination of them. With the condition that $c\le0$, it suffices to show that the first one, $(-2,1,0)$, cannot have a negative coefficient. This is equivalent to showing that $a+b+c\le0$, i.e. (substracting that from $a+2b+3c=0$) $b\ge-2c$. Assume $c\le0$ and let $C: =-c\ge0$. So the ineq is $M_1^{3C-2b}M_2^{2b}\ge M_3^{3C}$. if $b\leq \frac32C$, we get $LHS \le M_3^{3C-2b}M_3^{2b}=M_3^{3C}=RHS$, so all $x_i$ are equal, contradiction. Just missing the case $\frac32C\le b<2C.$ :