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  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $n \\ge 1$. For each $n$, the claim is the inequality for all $x_1, \\dots, x_n \\in I$ and all weights $w_i \\ge 0$ with sum $1$.\n\n1. First, a weighted average stays in $I$. Let $y = \\sum_{i=1}^{m} v_i x_i$ with $x_i \\in I$, $v_i \\ge 0$ and $\\sum_i v_i = 1$.\n2. Let $p = \\min_i x_i$ and $q = \\max_i x_i$. Then $p = \\sum_i v_i p \\le y \\le \\sum_i v_i q = q$.\n3. Both $p$ and $q$ lie in $I$, and $I$ is an [[interval|0TK8Y3J5RW]]. So $y \\in I$. In particular, both sides of the inequality are defined.\n4. Base case $n = 1$. Then $w_1 = 1$, and both sides equal $f(x_1)$.\n5. Induction step. Let $n \\ge 2$ and assume the claim for $n - 1$ points. Let $x_1, \\dots, x_n$ and $w_1, \\dots, w_n$ be as in the statement.\n6. If $w_n = 1$, then $w_i = 0$ for $i \u003c n$, since the weights are nonnegative with sum $1$. Both sides equal $f(x_n)$.\n7. Otherwise $w_n \u003c 1$. Put $s = 1 - w_n = w_1 + \\cdots + w_{n-1} \u003e 0$, and $y = \\sum_{i=1}^{n-1} \\frac{w_i}{s} x_i$.\n8. The weights $w_i / s$ for $i \u003c n$ are nonnegative with sum $1$. So $y \\in I$ by step 3.\n9. We have $\\sum_{i=1}^{n} w_i x_i = s y + w_n x_n$, with $s + w_n = 1$ and $s \\in (0, 1]$.\n10. By the definition of a [[convex function|0VV0HSW1ZS]] with $t = s$, $f(s y + w_n x_n) \\le s f(y) + w_n f(x_n)$.\n11. By the induction hypothesis for the $n - 1$ points $x_1, \\dots, x_{n-1}$, $f(y) \\le \\sum_{i=1}^{n-1} \\frac{w_i}{s} f(x_i)$.\n12. Multiplying by $s \u003e 0$ and using step 10 gives $f\\big(\\sum_{i=1}^{n} w_i x_i\\big) \\le \\sum_{i=1}^{n-1} w_i f(x_i) + w_n f(x_n) = \\sum_{i=1}^{n} w_i f(x_i)$.\n13. By induction, the inequality holds for every $n \\ge 1$.\n\n## Idea\n\nA weighted average of $n$ points is a two-point average of $x_n$ and the average $y$ of the other points. Convexity handles the two-point split, and the induction hypothesis handles $y$. The renormalized weights $w_i / s$ are what make the inner average a genuine average.\n",
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