My favorite example: it's not too hard to show---see here for example---that if $U$ is a unitary matrix, then $\left|\operatorname{Per}(U)\right|\le1$, where Per denotes the matrix permanent function,

$\operatorname{Per}(U) = \sum_{\sigma \in S_n} \prod_{i=1}^n u_{i,\sigma(i)}$.

However, one can also give an immediate "physics proof" of the same inequality, as follows. Given any $n\times n$ unitary matrix $U$, one can set up a quantum optics experiment where $n$ identical photons are generated in separate input ports and pass through a network of beamsplitters; then the total amplitude for a single photon to appear in each of $n$ output ports, with no "bunching" of multiple photons in the same port, is equal to $\operatorname{Per(U)}$. (Intuitively, this is because photons are bosons, so you need to sum over all $n!$ possible ways that they could be permuted, with each permutation contributing an amplitude that's a product of the transition amplitudes for the photons considered individually.)

OK, but the probability of a measurement outcome is just the squared absolute value of its amplitude, and probabilities can never exceed $1$. Therefore $\left|\operatorname{Per}(U)\right|\le1$.