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Let me prove it for $p=3$ (only because the notations look more friendly.) The maximum of left hand side is realized when the corresponding multiples are equally sorted, so let us assume that it is the case. I claim that not only sum, but $s$-th largest summand of RHS (clarification: first largest means maximal) is not less than in LHS, for any $s=1,2,\dots,n$. Denote by $\alpha, \beta, \gamma$ the $s$-th largest elements of the arrays $a, b, c$ respectively. Choose minimal indices $i, j, k$ respectively for which $a_i\geqslant \alpha$, $b_j\geqslant \beta$, $c_k\geqslant \gamma$. Without loss of generality $k=\max(i, j, k) $. Then $m_r\geqslant \alpha \beta \gamma$ for all $r$ such that $c_r\geqslant \gamma$ (because all such $r$ are not less than $\max(i, j)$), and we get at least $s$ such values of $r$, thus $s$-th largest $m$ is not less than $\alpha \beta\gamma$, as desired.