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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $A$ and $B$ be $n \\times n$ [[Hermitian matrices|0P17NGQQMQ]].\nThen\n$$\\tr e^{A + B} \\le \\tr\\left( e^A e^B \\right),$$\nwhere $e^X$ is the [[matrix exponential|020M04FN1H]] and $\\tr$ is the [[trace|0XC1RDBY7A]].\n\n## Remarks\n\nThe right side is a positive real number, because $\\tr(e^A e^B) = \\tr(e^{A/2} e^B e^{A/2})$ and $e^{A/2} e^B e^{A/2}$ is positive definite.\nOne proof uses the inequality $\\tr\\big((XY)^{2^k}\\big) \\le \\tr\\big(X^{2^k} Y^{2^k}\\big)$ for positive definite $X$ and $Y$, which follows from the Cauchy–Schwarz inequality for the trace by induction on $k$.\nPut $X = e^{A/2^k}$ and $Y = e^{B/2^k}$ and let $k \\to \\infty$; the [[Lie product formula|HK3BRMK912]] turns the left side into $\\tr e^{A + B}$.\nIf $AB = BA$, then $e^{A + B} = e^A e^B$ and equality holds.\nThe naive extension to three Hermitian matrices, with $\\abs{\\tr(e^A e^B e^C)}$ on the right, is false in general.\n\n## Explanations\n\n### L1 human\n\nFor numbers, $e^{a + b} = e^a e^b$.\nFor Hermitian matrices that do not commute, the two sides differ, but their traces compare in one direction: the trace of $e^{A + B}$ is at most the trace of $e^A e^B$.\nThis one-sided rule is often enough to bound quantities in physics and probability.\n\n### L2 human\n\nIn statistical mechanics, $\\tr e^{-\\beta H}$ is the partition function of a quantum system with Hamiltonian $H$, so the inequality bounds a partition function by a simpler expression.\nIn probability, it is used in the approach of Ahlswede and Winter to matrix concentration inequalities.\nThere one must control the trace of the exponential of a sum of random Hermitian matrices.\nLieb found a version for three matrices, with an integral in place of a simple product.\n",
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