Higher order version of Chebyshev's (algebraic) inequality

Question

Let $(a_{i})_{i=1}^{n}$ and $(b_{i})_{i=1}^{n}$ be finite monotone increasing sequences (or both mon. decreasing) of real numbers and $m_{i}>0$ with $\sum_{i=1}^{n}m_{i}=1$. Chebyshev's (algebraic) inequality tells us that \begin{equation} \sum_{i}a_{i}b_{i}m_{i}\ge\sum_{i}a_{i}m_{i}\sum_{i}b_{i}m_{i}, \end{equation} which follows easily by summing over $i,j$ in \begin{equation} m_{i}m_{j}(a_{i}-a_{j})(b_{i}-b_{j})\ge0. \end{equation} Now take one more increasing sequence $(c_{i})_{i=1}^{n}$ and assume $a,b,c$ are bounded in $[0,1]$. Is it true that \begin{equation} \sum_{i}a_{i}b_{i}m_{i}\sum_{i}a_{i}c_{i}m_{i}\sum_{i}b_{i}c_{i}m_{i}\ge\sum_{i}a_{i}m_{i}\sum_{i}b_{i}m_{i}\sum_{i}c_{i}m_{i}\sum_{i}a_{i}b_{i}c_{i}m_{i}? \end{equation} I'm happy about any answers, references,...