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  "body": "\n## Proof\n\n1. Let $x \u003e 0$, and let $y = \\ln x$, where $\\ln$ is the [[natural logarithm|0044CPHJBV]].\n2. By the inequality [[$1 + y \\le e^y$|012C7GFF20]], we get $1 + \\ln x \\le e^{\\ln x}$.\n3. By [[the inverse relation of $\\ln$ and $\\exp$|0RKW2X63YR]], $e^{\\ln x} = x$. So $1 + \\ln x \\le x$, that is, $\\ln x \\le x - 1$.\n4. The same inequality says that $1 + y = e^y$ holds if and only if $y = 0$. So $\\ln x = x - 1$ holds if and only if $\\ln x = 0$.\n5. If $\\ln x = 0$, then $x = e^{\\ln x} = e^0 = 1$. Here $e^0 = 1$ by the [[series of the exponential|0RMVS7RXZX]], since all terms after the first vanish at $0$.\n6. Conversely, if $x = 1$, then $\\ln x = \\ln(e^0) = 0$, since $\\ln(e^y) = y$ for every real $y$.\n7. So $\\ln x \\le x - 1$ for every $x \u003e 0$, with equality if and only if $x = 1$.\n\n## Idea\n\nThe line $y = 1 + t$ lies below the graph of $e^t$ and touches it only at $t = 0$. Substituting $t = \\ln x$ moves this fact over to the logarithm. The line $y = x - 1$ is the tangent to $\\ln$ at $x = 1$.\n",
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