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  "body": "\n## Statement\n\nLet $a, b, c$ be positive real numbers.\nThen\n$$\\frac{a}{b + c} + \\frac{b}{c + a} + \\frac{c}{a + b} \\ge \\frac{3}{2},$$\nwith equality if and only if $a = b = c$.\n\n## Remarks\n\nProof idea: let $s = a + b + c$.\nAdding $1$ to each fraction gives $\\frac{a}{b+c} + 1 = \\frac{s}{b+c}$, so the left side equals $s \\big( \\frac{1}{b+c} + \\frac{1}{c+a} + \\frac{1}{a+b} \\big) - 3$.\nBy the [[Cauchy–Schwarz inequality|0RFWW56Y22]], or the inequality between arithmetic and harmonic means, $\\big((b+c) + (c+a) + (a+b)\\big)\\big(\\frac{1}{b+c} + \\frac{1}{c+a} + \\frac{1}{a+b}\\big) \\ge 9$.\nSince the first factor is $2s$, the left side is at least $\\frac{9}{2} - 3 = \\frac{3}{2}$.\n\nExample: $a = b = 1$, $c = 2$ gives $\\frac{1}{3} + \\frac{1}{3} + 1 = \\frac{5}{3} \\ge \\frac{3}{2}$.\n\nThere is no upper bound: for $a = b = 1$ and $c \\to \\infty$ the left side tends to $\\infty$.\n\n## Explanations\n\n### L1 human\n\nDivide each of three positive numbers by the sum of the other two and add the results.\nYou always get at least $\\frac{3}{2}$, and you get exactly $\\frac{3}{2}$ only when the three numbers are equal.\n\n### L2 human\n\nNesbitt's inequality is a standard first example of a symmetric inequality in three variables.\nIt has many proofs: by the means inequalities, by the [[rearrangement inequality|6E9TAF8HCY]], or by clearing denominators and using [[Muirhead's inequality|PBPFVY20WT]].\nComparing the proofs is a good way to learn the toolbox of elementary inequalities.\n",
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