A homogenization principle for total variation We prove an inequality comparing the variational distance between pairs of product probability measures to its homogenized counterpart. If $P_1,\ldots,P_n,Q_1,\ldots,Q_n$ are arbitrary probability measures on a measurable space and $\bar P:=\frac1n\sum_{i=1}^n P_i, \bar Q:=\frac1n\sum_{i=1}^n Q_i $, we show that $$TV\!\left(\bigotimes_{i=1}^n P_i, \bigotimes_{i=1}^n Q_i\right) \;\ge\; c\,TV(\bar P^{\otimes n},\bar Q^{\otimes n}),$$ where $c>0$ is a universal constant. The proof is based on a one-dimensional representation of total variation between products. We embed pairs of probability distributions $P_i,Q_i$ into positive measures $η_i$ on $\mathbb{R}$. We then define a functional $T$ over measures on $\mathbb{R}$ that realizes TV over products via convolution: $TV\!\left(\bigotimes_{i=1}^n P_i, \bigotimes_{i=1}^n Q_i\right)=T(η_1*\cdots *η_n)$. Our main analytic discovery is that for the relevant class of positive measures $η_i$, the convolution inequality $T(η_1*\cdots*η_n) \ge c\,T\!\left(\barη^{*n}\right)$ holds, where $\barη=\frac1n\sum_{i=1}^n η_i$. Finally, a higher-dimensional lifting argument shows that $T\!\left(\barη^{*n}\right)\ge TV(\bar P^{\otimes n},\bar Q^{\otimes n})$. To our knowledge, both the exact representation and the convolution inequality are new.