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  "body": "\n## Statement\n\nLet $I$ be an [[interval|0TK8Y3J5RW]] of $\\R$ and let $f \\colon I \\to \\R$ be a [[convex function|0VV0HSW1ZS]].\nLet $x_1 \\ge x_2 \\ge \\dots \\ge x_n$ and $y_1 \\ge y_2 \\ge \\dots \\ge y_n$ be numbers in $I$ such that $(x_1, \\dots, x_n)$ [[majorizes|5K7VT2VKZT]] $(y_1, \\dots, y_n)$, that is,\n$$\\sum_{i=1}^{k} x_i \\ge \\sum_{i=1}^{k} y_i \\quad \\text{for } k = 1, \\dots, n - 1, \\qquad \\sum_{i=1}^{n} x_i = \\sum_{i=1}^{n} y_i.$$\nThen\n$$\\sum_{i=1}^{n} f(x_i) \\ge \\sum_{i=1}^{n} f(y_i).$$\n\n## Remarks\n\nProof idea: let $c_i$ be the slope of the chord of $f$ between $x_i$ and $y_i$, or a [[subderivative|9C5WXYYQEK]] at $x_i$ when $x_i = y_i$.\nSince [[the slopes of a convex function increase|0BY5FTHEEE]] and both sequences decrease, $c_1 \\ge c_2 \\ge \\dots \\ge c_n$.\nThen $\\sum_i \\big(f(x_i) - f(y_i)\\big) = \\sum_i c_i (x_i - y_i)$, and summation by parts writes this as a sum of products of the nonnegative numbers $c_k - c_{k+1}$ with the nonnegative partial sums $\\sum_{i \\le k} (x_i - y_i)$.\n\nExample: $(2, 0)$ majorizes $(1, 1)$, so $f(2) + f(0) \\ge 2 f(1)$, which is midpoint convexity.\nMore generally $(x_1, \\dots, x_n)$ majorizes the constant vector $(\\bar x, \\dots, \\bar x)$ with $\\bar x$ the mean, which gives [[Jensen's inequality|0RWMRTPC87]] with equal weights.\n\nThe equality of the total sums is needed.\nWith $n = 1$, $f(x) = -x$, $x_1 = 1$ and $y_1 = 0$, one has $x_1 \\ge y_1$ but $f(x_1) = -1 \u003c 0 = f(y_1)$.\n\n## Explanations\n\n### L1 human\n\nKaramata's inequality compares two lists of numbers with the same total.\nIf one list is more spread out than the other, then summing a convex function over it gives a larger result.\nA convex function rewards spreading out, because its graph bends upward.\n\n### L2 human\n\nThe inequality says that $x \\mapsto \\sum_i f(x_i)$ is a [[Schur-convex function|A72XTVF2GK]] whenever $f$ is convex.\nIt contains Jensen's inequality with equal weights and implies many symmetric inequalities, for instance by comparing a vector with a less spread out one.\nA converse also holds: if the conclusion is true for every convex $f$, then $x$ majorizes $y$.\n",
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