{
  "index": "1991-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $a$ and $b$ be positive numbers. Find the largest number $c$, in\nterms of $a$ and $b$, such that\n\\[\na^x b^{1-x} \\leq a \\frac{\\sinh ux}{\\sinh u} + b \\frac{\\sinh u(1-x)}{\\sinh u}\n\\]\nfor all $u$ with $0 < |u| \\leq c$ and for all $x$, $0 < x < 1$. (Note:\n$\\sinh u = (e^u - e^{-u})/2$.)\n\\end{itemize}\n\\end{document}",
  "solution": "Solution 1. We will show that \\( c=|\\ln (a / b)| \\) by proving that the inequality is satisfied if and only if \\( 0<|u| \\leq|\\ln (a / b)| \\). The right-hand side is an even function of \\( u \\); hence it suffices to consider \\( u>0 \\). Replacing \\( x \\) by \\( 1-x \\) and interchanging \\( a \\) and \\( b \\) preserves the inequality, so we may assume \\( a \\geq b \\).\n\nWe will show that for each \\( x \\in(0,1) \\), the function\n\\[\nF(u)=a \\frac{\\sinh u x}{\\sinh u}+b \\frac{\\sinh u(1-x)}{\\sinh u}-a^{x} b^{1-x}\n\\]\nof \\( u \\) is decreasing for \\( u>0 \\). For this, it suffices to show that for each \\( \\alpha \\in(0,1) \\), the function\n\\[\nf(u)=\\frac{\\sinh \\alpha u}{\\sinh u}\n\\]\nis decreasing for \\( u>0 \\), since \\( F(u) \\) is a constant plus a sum of positive multiples of two such functions \\( f \\). Differentiating with respect to \\( u \\) yields\n\\[\nf^{\\prime}(u)=\\frac{\\alpha \\cosh (\\alpha u) \\sinh u-\\sinh (\\alpha u) \\cosh u}{\\sinh ^{2} u}\n\\]\nwhich for \\( u>0 \\) is negative if and only if the value of the function\n\\[\ng(u)=\\alpha \\tanh u-\\tanh (\\alpha u)\n\\]\nis negative. The negativity of \\( g(u) \\) for \\( u>0 \\) follows from \\( g(0)=0 \\) and the negativity of\n\\[\ng^{\\prime}(u)=\\frac{\\alpha}{\\cosh ^{2} u}-\\frac{\\alpha}{\\cosh ^{2} \\alpha u}\n\\]\nwhich in turn follows since cosh is increasing on \\( \\mathbb{R}^{+} \\). Thus \\( F(u) \\) is decreasing for \\( u>0 \\).\n\nIf \\( a>b \\) then \\( F(u) \\) is zero at \\( u=\\ln (a / b) \\). If \\( a=b \\) then \\( \\lim _{u \\rightarrow 0^{+}} F(u)=0 \\) by L'Hôpital's Rule [Spv, Ch. 11, Theorem 9]. Since \\( F(u) \\) is decreasing for \\( u>0 \\), in both cases we have \\( F(u) \\geq 0 \\) for \\( 0<u \\leq \\ln (a / b) \\) and \\( F(u)<0 \\) for \\( u>\\ln (a / b) \\), as desired.\n\nSolution 2. As before, we may assume \\( u>0 \\) and \\( a \\geq b \\). Taking \\( \\ln \\) of each side and rearranging, we find that the inequality is equivalent to\n\\[\nx(\\ln (a \\sinh u))+(1-x)(\\ln (b \\sinh u)) \\leq \\ln (a \\sinh u x+b \\sinh u(1-x)),\n\\]\nwhich, if we define\n\\[\nf(x)=a \\sinh u x+b \\sinh u(1-x),\n\\]\ncan be written as\n\\[\nx \\ln f(1)+(1-x) \\ln f(0) \\leq \\ln f(x)\n\\]\n\nThis will hold if \\( u \\) is such that \\( \\ln f(x) \\) is concave for \\( x \\in(0,1) \\), and will fail if \\( \\ln f(x) \\) is strictly convex on \\( (0,1) \\).\n\nWe compute\n\\[\n\\begin{aligned}\n\\frac{d}{d x} \\ln f= & \\frac{f^{\\prime}}{f} \\\\\n\\frac{d^{2}}{d x^{2}} \\ln f= & \\frac{f f^{\\prime \\prime}-\\left(f^{\\prime}\\right)^{2}}{f^{2}} \\\\\nf^{\\prime}= & a u \\cosh u x-b u \\cosh u(1-x) \\\\\nf^{\\prime \\prime}= & a u^{2} \\sinh u x+b u^{2} \\sinh u(1-x) \\\\\nf f^{\\prime \\prime}-\\left(f^{\\prime}\\right)^{2}= & \\left(a^{2}+b^{2}\\right) u^{2}\\left(\\sinh ^{2} u x-\\cosh ^{2} u x\\right) \\\\\n& +2 a b u^{2}(\\sinh u x \\sinh u(1-x)+\\cosh u x \\cosh u(1-x)) \\\\\n= & \\left(a^{2}+b^{2}\\right) u^{2}(-1)+2 a b u^{2} \\cosh (u x+u(1-x)) \\\\\n= & u^{2}\\left(-a^{2}-b^{2}+2 a b \\cosh u\\right)\n\\end{aligned}\n\\]\no \\( \\frac{d^{2}}{d x^{2}} \\ln f \\) has constant sign for \\( x \\in(0,1) \\), and the following are equivalent:\n\\[\n\\begin{aligned}\n\\frac{d^{2}}{d x^{2}} \\ln f & \\leq 0 \\quad \\text { for } x \\in(0,1) \\\\\n-a^{2}-b^{2}+2 a b \\cosh u & \\leq 0 \\\\\n-a^{2}-b^{2}+a b e^{u}+a b e^{-u} & \\leq 0 \\\\\n-\\left(a-e^{u} b\\right)\\left(a-e^{-u} b\\right) & \\leq 0 \\\\\n-\\left(a-e^{u} b\\right) & \\leq 0 \\quad(\\text { since } u>0 \\text { and } a \\geq b>0) \\\\\nu & \\leq \\ln (a / b) .\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "u",
    "F",
    "f",
    "g",
    "\\\\alpha"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "firstpos",
        "b": "secondpos",
        "c": "largestc",
        "x": "mixratio",
        "u": "shiftvar",
        "F": "mainfun",
        "f": "smallfun",
        "g": "auxifun",
        "\\alpha": "alphavar"
      },
      "question": "Let $firstpos$ and $secondpos$ be positive numbers. Find the largest number $largestc$, in terms of $firstpos$ and $secondpos$, such that\n\\[\nfirstpos^{mixratio}\\, secondpos^{1-mixratio} \\leq firstpos \\frac{\\sinh{\\,shiftvar mixratio}}{\\sinh{\\,shiftvar}} + secondpos \\frac{\\sinh{\\,shiftvar(1-mixratio)}}{\\sinh{\\,shiftvar}}\n\\]\nfor all $shiftvar$ with $0 < |shiftvar| \\leq largestc$ and for all $mixratio$, $0 < mixratio < 1$. (Note: $\\sinh shiftvar = (e^{shiftvar} - e^{-shiftvar})/2$.)",
      "solution": "Solution 1. We will show that \\( largestc=|\\ln (firstpos / secondpos)| \\) by proving that the inequality is satisfied if and only if \\( 0<|shiftvar| \\leq|\\ln (firstpos / secondpos)| \\). The right-hand side is an even function of \\( shiftvar \\); hence it suffices to consider \\( shiftvar>0 \\). Replacing \\( mixratio \\) by \\( 1-mixratio \\) and interchanging \\( firstpos \\) and \\( secondpos \\) preserves the inequality, so we may assume \\( firstpos \\geq secondpos \\).\n\nWe will show that for each \\( mixratio \\in(0,1) \\), the function\n\\[\nmainfun(shiftvar)=firstpos \\frac{\\sinh shiftvar mixratio}{\\sinh shiftvar}+secondpos \\frac{\\sinh shiftvar(1-mixratio)}{\\sinh shiftvar}-firstpos^{mixratio} secondpos^{1-mixratio}\n\\]\nof \\( shiftvar \\) is decreasing for \\( shiftvar>0 \\). For this, it suffices to show that for each \\( alphavar \\in(0,1) \\), the function\n\\[\nsmallfun(shiftvar)=\\frac{\\sinh alphavar \\, shiftvar}{\\sinh shiftvar}\n\\]\nis decreasing for \\( shiftvar>0 \\), since \\( mainfun(shiftvar) \\) is a constant plus a sum of positive multiples of two such functions \\( smallfun \\). Differentiating with respect to \\( shiftvar \\) yields\n\\[\nsmallfun^{\\prime}(shiftvar)=\\frac{alphavar \\cosh (alphavar \\, shiftvar) \\sinh shiftvar-\\sinh (alphavar \\, shiftvar) \\cosh shiftvar}{\\sinh ^{2} shiftvar}\n\\]\nwhich for \\( shiftvar>0 \\) is negative if and only if the value of the function\n\\[\nauxifun(shiftvar)=alphavar \\tanh shiftvar-\\tanh (alphavar \\, shiftvar)\n\\]\nis negative. The negativity of \\( auxifun(shiftvar) \\) for \\( shiftvar>0 \\) follows from \\( auxifun(0)=0 \\) and the negativity of\n\\[\nauxifun^{\\prime}(shiftvar)=\\frac{alphavar}{\\cosh ^{2} shiftvar}-\\frac{alphavar}{\\cosh ^{2} alphavar \\, shiftvar}\n\\]\nwhich in turn follows since cosh is increasing on \\( \\mathbb{R}^{+} \\). Thus \\( mainfun(shiftvar) \\) is decreasing for \\( shiftvar>0 \\).\n\nIf \\( firstpos>secondpos \\) then \\( mainfun(shiftvar) \\) is zero at \\( shiftvar=\\ln (firstpos / secondpos) \\). If \\( firstpos=secondpos \\) then \\( \\lim _{shiftvar \\rightarrow 0^{+}} mainfun(shiftvar)=0 \\) by L'Hôpital's Rule [Spv, Ch. 11, Theorem 9]. Since \\( mainfun(shiftvar) \\) is decreasing for \\( shiftvar>0 \\), in both cases we have \\( mainfun(shiftvar) \\geq 0 \\) for \\( 0<shiftvar \\leq \\ln (firstpos / secondpos) \\) and \\( mainfun(shiftvar)<0 \\) for \\( shiftvar>\\ln (firstpos / secondpos) \\), as desired.\n\nSolution 2. As before, we may assume \\( shiftvar>0 \\) and \\( firstpos \\geq secondpos \\). Taking \\( \\ln \\) of each side and rearranging, we find that the inequality is equivalent to\n\\[\nmixratio(\\ln (firstpos \\sinh shiftvar))+(1-mixratio)(\\ln (secondpos \\sinh shiftvar)) \\leq \\ln (firstpos \\sinh shiftvar \\, mixratio+secondpos \\sinh shiftvar(1-mixratio)),\n\\]\nwhich, if we define\n\\[\nsmallfun(mixratio)=firstpos \\sinh shiftvar \\, mixratio+secondpos \\sinh shiftvar(1-mixratio),\n\\]\ncan be written as\n\\[\nmixratio \\ln smallfun(1)+(1-mixratio) \\ln smallfun(0) \\leq \\ln smallfun(mixratio)\n\\]\n\nThis will hold if \\( shiftvar \\) is such that \\( \\ln smallfun(mixratio) \\) is concave for \\( mixratio \\in(0,1) \\), and will fail if \\( \\ln smallfun(mixratio) \\) is strictly convex on \\( (0,1) \\).\n\nWe compute\n\\[\n\\begin{aligned}\n\\frac{d}{d mixratio} \\ln smallfun= & \\frac{smallfun^{\\prime}}{smallfun} \\\\\n\\frac{d^{2}}{d mixratio^{2}} \\ln smallfun= & \\frac{smallfun \\, smallfun^{\\prime \\prime}-\\left(smallfun^{\\prime}\\right)^{2}}{smallfun^{2}} \\\\\nsmallfun^{\\prime}= & firstpos \\, shiftvar \\cosh shiftvar \\, mixratio-secondpos \\, shiftvar \\cosh shiftvar(1-mixratio) \\\\\nsmallfun^{\\prime \\prime}= & firstpos \\, shiftvar^{2} \\sinh shiftvar \\, mixratio+secondpos \\, shiftvar^{2} \\sinh shiftvar(1-mixratio) \\\\\nsmallfun \\, smallfun^{\\prime \\prime}-\\left(smallfun^{\\prime}\\right)^{2}= & \\left(firstpos^{2}+secondpos^{2}\\right) shiftvar^{2}\\left(\\sinh ^{2} shiftvar \\, mixratio-\\cosh ^{2} shiftvar \\, mixratio\\right) \\\\\n& +2 firstpos \\, secondpos \\, shiftvar^{2}(\\sinh shiftvar \\, mixratio \\sinh shiftvar(1-mixratio)+\\cosh shiftvar \\, mixratio \\cosh shiftvar(1-mixratio)) \\\\\n= & \\left(firstpos^{2}+secondpos^{2}\\right) shiftvar^{2}(-1)+2 firstpos \\, secondpos \\, shiftvar^{2} \\cosh (shiftvar \\, mixratio+shiftvar(1-mixratio)) \\\\\n= & shiftvar^{2}\\left(-firstpos^{2}-secondpos^{2}+2 firstpos \\, secondpos \\cosh shiftvar\\right)\n\\end{aligned}\n\\]\nso \\( \\frac{d^{2}}{d mixratio^{2}} \\ln smallfun \\) has constant sign for \\( mixratio \\in(0,1) \\), and the following are equivalent:\n\\[\n\\begin{aligned}\n\\frac{d^{2}}{d mixratio^{2}} \\ln smallfun & \\leq 0 \\quad \\text { for } mixratio \\in(0,1) \\\\\n-firstpos^{2}-secondpos^{2}+2 firstpos \\, secondpos \\cosh shiftvar & \\leq 0 \\\\\n-firstpos^{2}-secondpos^{2}+firstpos \\, secondpos e^{shiftvar}+firstpos \\, secondpos e^{-shiftvar} & \\leq 0 \\\\\n-\\left(firstpos-e^{shiftvar} secondpos\\right)\\left(firstpos-e^{-shiftvar} secondpos\\right) & \\leq 0 \\\\\n-\\left(firstpos-e^{shiftvar} secondpos\\right) & \\leq 0 \\quad(\\text { since } shiftvar>0 \\text { and } firstpos \\geq secondpos>0) \\\\\nshiftvar & \\leq \\ln (firstpos / secondpos) .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "honeycomb",
        "u": "butterscotch",
        "F": "raspberries",
        "f": "marshmallow",
        "g": "cheeseball",
        "\\alpha": "shipwreck",
        "a": "lighthouse",
        "b": "snowflake",
        "c": "driftwood"
      },
      "question": "Let $lighthouse$ and $snowflake$ be positive numbers. Find the largest number $driftwood$, in\nterms of $lighthouse$ and $snowflake$, such that\n\\[\nlighthouse^{honeycomb} snowflake^{1-honeycomb} \\leq lighthouse \\frac{\\sinh butterscotch\\, honeycomb}{\\sinh butterscotch} + snowflake \\frac{\\sinh butterscotch(1-honeycomb)}{\\sinh butterscotch}\n\\]\nfor all $butterscotch$ with $0 < |butterscotch| \\leq driftwood$ and for all $honeycomb$, $0 < honeycomb < 1$. (Note:\n$\\sinh butterscotch = (e^{butterscotch} - e^{-butterscotch})/2$.)\n\\end{itemize}\n\\end{document}",
      "solution": "Solution 1. We will show that \\( driftwood=|\\ln (lighthouse / snowflake)| \\) by proving that the inequality is satisfied if and only if \\( 0<|butterscotch| \\leq|\\ln (lighthouse / snowflake)| \\). The right-hand side is an even function of \\( butterscotch \\); hence it suffices to consider \\( butterscotch>0 \\). Replacing \\( honeycomb \\) by \\( 1-honeycomb \\) and interchanging \\( lighthouse \\) and \\( snowflake \\) preserves the inequality, so we may assume \\( lighthouse \\geq snowflake \\).\n\nWe will show that for each \\( honeycomb \\in(0,1) \\), the function\n\\[\nraspberries(butterscotch)=lighthouse \\frac{\\sinh butterscotch\\, honeycomb}{\\sinh butterscotch}+snowflake \\frac{\\sinh butterscotch(1-honeycomb)}{\\sinh butterscotch}-lighthouse^{honeycomb} snowflake^{1-honeycomb}\n\\]\nof \\( butterscotch \\) is decreasing for \\( butterscotch>0 \\). For this, it suffices to show that for each \\( shipwreck \\in(0,1) \\), the function\n\\[\nmarshmallow(butterscotch)=\\frac{\\sinh shipwreck\\, butterscotch}{\\sinh butterscotch}\n\\]\nis decreasing for \\( butterscotch>0 \\), since \\( raspberries(butterscotch) \\) is a constant plus a sum of positive multiples of two such functions \\( marshmallow \\). Differentiating with respect to \\( butterscotch \\) yields\n\\[\nmarshmallow^{\\prime}(butterscotch)=\\frac{shipwreck \\cosh (shipwreck\\, butterscotch) \\sinh butterscotch-\\sinh (shipwreck\\, butterscotch) \\cosh butterscotch}{\\sinh ^{2} butterscotch}\n\\]\nwhich for \\( butterscotch>0 \\) is negative if and only if the value of the function\n\\[\ncheeseball(butterscotch)=shipwreck \\tanh butterscotch-\\tanh (shipwreck\\, butterscotch)\n\\]\nis negative. The negativity of \\( cheeseball(butterscotch) \\) for \\( butterscotch>0 \\) follows from \\( cheeseball(0)=0 \\) and the negativity of\n\\[\ncheeseball^{\\prime}(butterscotch)=\\frac{shipwreck}{\\cosh ^{2} butterscotch}-\\frac{shipwreck}{\\cosh ^{2} shipwreck\\, butterscotch}\n\\]\nwhich in turn follows since cosh is increasing on \\( \\mathbb{R}^{+} \\). Thus \\( raspberries(butterscotch) \\) is decreasing for \\( butterscotch>0 \\).\n\nIf \\( lighthouse>snowflake \\) then \\( raspberries(butterscotch) \\) is zero at \\( butterscotch=\\ln (lighthouse / snowflake) \\). If \\( lighthouse=snowflake \\) then \\( \\lim _{butterscotch \\rightarrow 0^{+}} raspberries(butterscotch)=0 \\) by L'H\\^opital's Rule [Spv, Ch. 11, Theorem 9]. Since \\( raspberries(butterscotch) \\) is decreasing for \\( butterscotch>0 \\), in both cases we have \\( raspberries(butterscotch) \\geq 0 \\) for \\( 0<butterscotch \\leq \\ln (lighthouse / snowflake) \\) and \\( raspberries(butterscotch)<0 \\) for \\( butterscotch>\\ln (lighthouse / snowflake) \\), as desired.\n\nSolution 2. As before, we may assume \\( butterscotch>0 \\) and \\( lighthouse \\geq snowflake \\). Taking \\( \\ln \\) of each side and rearranging, we find that the inequality is equivalent to\n\\[\nhoneycomb(\\ln (lighthouse \\sinh butterscotch))+(1-honeycomb)(\\ln (snowflake \\sinh butterscotch)) \\leq \\ln (lighthouse \\sinh butterscotch\\, honeycomb+snowflake \\sinh butterscotch(1-honeycomb)),\n\\]\nwhich, if we define\n\\[\nmarshmallow(honeycomb)=lighthouse \\sinh butterscotch\\, honeycomb+snowflake \\sinh butterscotch(1-honeycomb),\n\\]\ncan be written as\n\\[\nhoneycomb \\ln marshmallow(1)+(1-honeycomb) \\ln marshmallow(0) \\leq \\ln marshmallow(honeycomb)\n\\]\n\nThis will hold if \\( butterscotch \\) is such that \\( \\ln marshmallow(honeycomb) \\) is concave for \\( honeycomb \\in(0,1) \\), and will fail if \\( \\ln marshmallow(honeycomb) \\) is strictly convex on \\( (0,1) \\).\n\nWe compute\n\\[\n\\begin{aligned}\n\\frac{d}{d honeycomb} \\ln marshmallow=& \\frac{marshmallow^{\\prime}}{marshmallow} \\\\\n\\frac{d^{2}}{d honeycomb^{2}} \\ln marshmallow=& \\frac{marshmallow \\, marshmallow^{\\prime \\prime}-\\left(marshmallow^{\\prime}\\right)^{2}}{marshmallow^{2}} \\\\\nmarshmallow^{\\prime}= & lighthouse \\, butterscotch \\cosh butterscotch\\, honeycomb-snowflake \\, butterscotch \\cosh butterscotch(1-honeycomb) \\\\\nmarshmallow^{\\prime \\prime}= & lighthouse \\, butterscotch^{2} \\sinh butterscotch\\, honeycomb+snowflake \\, butterscotch^{2} \\sinh butterscotch(1-honeycomb) \\\\\nmarshmallow \\, marshmallow^{\\prime \\prime}-\\left(marshmallow^{\\prime}\\right)^{2}= & \\left(lighthouse^{2}+snowflake^{2}\\right) butterscotch^{2}\\left(\\sinh ^{2} butterscotch\\, honeycomb-\\cosh ^{2} butterscotch\\, honeycomb\\right) \\\\\n& +2 lighthouse \\, snowflake \\, butterscotch^{2}(\\sinh butterscotch\\, honeycomb \\sinh butterscotch(1-honeycomb)+\\cosh butterscotch\\, honeycomb \\cosh butterscotch(1-honeycomb)) \\\\\n= & \\left(lighthouse^{2}+snowflake^{2}\\right) butterscotch^{2}(-1)+2 lighthouse \\, snowflake \\, butterscotch^{2} \\cosh (butterscotch\\, honeycomb+butterscotch(1-honeycomb)) \\\\\n= & butterscotch^{2}\\left(-lighthouse^{2}-snowflake^{2}+2 lighthouse \\, snowflake \\cosh butterscotch\\right)\n\\end{aligned}\n\\]\nso \\( \\frac{d^{2}}{d honeycomb^{2}} \\ln marshmallow \\) has constant sign for \\( honeycomb \\in(0,1) \\), and the following are equivalent:\n\\[\n\\begin{aligned}\n\\frac{d^{2}}{d honeycomb^{2}} \\ln marshmallow & \\leq 0 \\quad \\text { for } honeycomb \\in(0,1) \\\\\n-lighthouse^{2}-snowflake^{2}+2 lighthouse \\, snowflake \\cosh butterscotch & \\leq 0 \\\\\n-lighthouse^{2}-snowflake^{2}+lighthouse \\, snowflake e^{butterscotch}+lighthouse \\, snowflake e^{-butterscotch} & \\leq 0 \\\\\n-\\left(lighthouse-e^{butterscotch} snowflake\\right)\\left(lighthouse-e^{-butterscotch} snowflake\\right) & \\leq 0 \\\\\n-\\left(lighthouse-e^{butterscotch} snowflake\\right) & \\leq 0 \\quad(\\text { since } butterscotch>0 \\text { and } lighthouse \\geq snowflake>0) \\\\\nbutterscotch & \\leq \\ln (lighthouse / snowflake) .\n\\end{aligned}\n\\]\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "u": "stationary",
        "F": "entropyfun",
        "f": "disorder",
        "g": "chaosmeter",
        "\\\\alpha": "omegafactor",
        "a": "negativeone",
        "b": "negativetwo",
        "c": "smallest"
      },
      "question": "Let $negativeone$ and $negativetwo$ be positive numbers. Find the largest number $smallest$, in\nterms of $negativeone$ and $negativetwo$, such that\n\\[\nnegativeone^{constantval} negativetwo^{1-constantval} \\leq negativeone \\frac{\\sinh stationary constantval}{\\sinh stationary} + negativetwo \\frac{\\sinh stationary(1-constantval)}{\\sinh stationary}\n\\]\nfor all $stationary$ with $0 < |stationary| \\leq smallest$ and for all $constantval$, $0 < constantval < 1$. (Note:\n$\\sinh stationary = (e^{stationary} - e^{-stationary})/2$.)\n\\end{itemize}\n\\end{document}",
      "solution": "Solution 1. We will show that \\( smallest=|\\ln (negativeone / negativetwo)| \\) by proving that the inequality is satisfied if and only if \\( 0<|stationary| \\leq|\\ln (negativeone / negativetwo)| \\). The right-hand side is an even function of \\( stationary \\); hence it suffices to consider \\( stationary>0 \\). Replacing \\( constantval \\) by \\( 1-constantval \\) and interchanging \\( negativeone \\) and \\( negativetwo \\) preserves the inequality, so we may assume \\( negativeone \\geq negativetwo \\).\n\nWe will show that for each \\( constantval \\in(0,1) \\), the function\n\\[\nentropyfun(stationary)=negativeone \\frac{\\sinh stationary constantval}{\\sinh stationary}+negativetwo \\frac{\\sinh stationary(1-constantval)}{\\sinh stationary}-negativeone^{constantval} negativetwo^{1-constantval}\n\\]\nof \\( stationary \\) is decreasing for \\( stationary>0 \\). For this, it suffices to show that for each \\( omegafactor \\in(0,1) \\), the function\n\\[\ndisorder(stationary)=\\frac{\\sinh omegafactor stationary}{\\sinh stationary}\n\\]\nis decreasing for \\( stationary>0 \\), since \\( entropyfun(stationary) \\) is a constant plus a sum of positive multiples of two such functions \\( disorder \\). Differentiating with respect to \\( stationary \\) yields\n\\[\ndisorder^{\\prime}(stationary)=\\frac{omegafactor \\cosh (omegafactor stationary) \\sinh stationary-\\sinh (omegafactor stationary) \\cosh stationary}{\\sinh ^{2} stationary}\n\\]\nwhich for \\( stationary>0 \\) is negative if and only if the value of the function\n\\[\nchaosmeter(stationary)=omegafactor \\tanh stationary-\\tanh (omegafactor stationary)\n\\]\nis negative. The negativity of \\( chaosmeter(stationary) \\) for \\( stationary>0 \\) follows from \\( chaosmeter(0)=0 \\) and the negativity of\n\\[\nchaosmeter^{\\prime}(stationary)=\\frac{omegafactor}{\\cosh ^{2} stationary}-\\frac{omegafactor}{\\cosh ^{2} omegafactor stationary}\n\\]\nwhich in turn follows since cosh is increasing on \\( \\mathbb{R}^{+} \\). Thus \\( entropyfun(stationary) \\) is decreasing for \\( stationary>0 \\).\n\nIf \\( negativeone>negativetwo \\) then \\( entropyfun(stationary) \\) is zero at \\( stationary=\\ln (negativeone / negativetwo) \\). If \\( negativeone=negativetwo \\) then \\( \\lim _{stationary \\rightarrow 0^{+}} entropyfun(stationary)=0 \\) by L'Hôpital's Rule [Spv, Ch. 11, Theorem 9]. Since \\( entropyfun(stationary) \\) is decreasing for \\( stationary>0 \\), in both cases we have \\( entropyfun(stationary) \\geq 0 \\) for \\( 0<stationary \\leq \\ln (negativeone / negativetwo) \\) and \\( entropyfun(stationary)<0 \\) for \\( stationary>\\ln (negativeone / negativetwo) \\), as desired.\n\nSolution 2. As before, we may assume \\( stationary>0 \\) and \\( negativeone \\geq negativetwo \\). Taking \\( \\ln \\) of each side and rearranging, we find that the inequality is equivalent to\n\\[\nconstantval(\\ln (negativeone \\sinh stationary))+(1-constantval)(\\ln (negativetwo \\sinh stationary)) \\leq \\ln (negativeone \\sinh stationary constantval+negativetwo \\sinh stationary(1-constantval)),\n\\]\nwhich, if we define\n\\[\ndisorder(constantval)=negativeone \\sinh stationary constantval+negativetwo \\sinh stationary(1-constantval),\n\\]\ncan be written as\n\\[\nconstantval \\ln disorder(1)+(1-constantval) \\ln disorder(0) \\leq \\ln disorder(constantval)\n\\]\n\nThis will hold if \\( stationary \\) is such that \\( \\ln disorder(constantval) \\) is concave for \\( constantval \\in(0,1) \\), and will fail if \\( \\ln disorder(constantval) \\) is strictly convex on \\( (0,1) \\).\n\nWe compute\n\\[\n\\begin{aligned}\n\\frac{d}{d constantval} \\ln disorder=& \\frac{disorder^{\\prime}}{disorder} \\\\\n\\frac{d^{2}}{d constantval^{2}} \\ln disorder=& \\frac{disorder disorder^{\\prime \\prime}-\\left(disorder^{\\prime}\\right)^{2}}{disorder^{2}} \\\\\ndisorder^{\\prime}=& negativeone stationary \\cosh stationary constantval-negativetwo stationary \\cosh stationary(1-constantval) \\\\\ndisorder^{\\prime \\prime}=& negativeone stationary^{2} \\sinh stationary constantval+negativetwo stationary^{2} \\sinh stationary(1-constantval) \\\\\ndisorder disorder^{\\prime \\prime}-\\left(disorder^{\\prime}\\right)^{2}=& \\left(negativeone^{2}+negativetwo^{2}\\right) stationary^{2}\\left(\\sinh ^{2} stationary constantval-\\cosh ^{2} stationary constantval\\right) \\\\\n& +2 negativeone negativetwo stationary^{2}(\\sinh stationary constantval \\sinh stationary(1-constantval)+\\cosh stationary constantval \\cosh stationary(1-constantval)) \\\\\n=& \\left(negativeone^{2}+negativetwo^{2}\\right) stationary^{2}(-1)+2 negativeone negativetwo stationary^{2} \\cosh (stationary constantval+stationary(1-constantval)) \\\\\n=& stationary^{2}\\left(-negativeone^{2}-negativetwo^{2}+2 negativeone negativetwo \\cosh stationary\\right)\n\\end{aligned}\n\\]\no \\( \\frac{d^{2}}{d constantval^{2}} \\ln disorder \\) has constant sign for \\( constantval \\in(0,1) \\), and the following are equivalent:\n\\[\n\\begin{aligned}\n\\frac{d^{2}}{d constantval^{2}} \\ln disorder & \\leq 0 \\quad \\text { for } constantval \\in(0,1) \\\\\n-negativeone^{2}-negativetwo^{2}+2 negativeone negativetwo \\cosh stationary & \\leq 0 \\\\\n-negativeone^{2}-negativetwo^{2}+negativeone negativetwo e^{stationary}+negativeone negativetwo e^{-stationary} & \\leq 0 \\\\\n-\\left(negativeone-e^{stationary} negativetwo\\right)\\left(negativeone-e^{-stationary} negativetwo\\right) & \\leq 0 \\\\\n-\\left(negativeone-e^{stationary} negativetwo\\right) & \\leq 0 \\quad(\\text { since } stationary>0 \\text { and } negativeone \\geq negativetwo>0) \\\\\nstationary & \\leq \\ln (negativeone / negativetwo) .\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "vbnmtrwe",
        "x": "mnbvcxzl",
        "u": "asdfghjk",
        "F": "poiulkjh",
        "f": "qazxswed",
        "g": "rfvbgtky",
        "\\alpha": "\\plmoknij"
      },
      "question": "Let $qzxwvtnp$ and $hjgrksla$ be positive numbers. Find the largest number $vbnmtrwe$, in\nterms of $qzxwvtnp$ and $hjgrksla$, such that\n\\[\nqzxwvtnp^{mnbvcxzl} hjgrksla^{1-mnbvcxzl} \\leq qzxwvtnp \\frac{\\sinh asdfghjk mnbvcxzl}{\\sinh asdfghjk} + hjgrksla \\frac{\\sinh asdfghjk(1-mnbvcxzl)}{\\sinh asdfghjk}\n\\]\nfor all $asdfghjk$ with $0 < |asdfghjk| \\leq vbnmtrwe$ and for all $mnbvcxzl$, $0 < mnbvcxzl < 1$. (Note:\n$\\sinh asdfghjk = (e^{asdfghjk} - e^{-asdfghjk})/2$.)",
      "solution": "Solution 1. We will show that \\( vbnmtrwe=|\\ln (qzxwvtnp / hjgrksla)| \\) by proving that the inequality is satisfied if and only if \\( 0<|asdfghjk| \\leq|\\ln (qzxwvtnp / hjgrksla)| \\). The right-hand side is an even function of \\( asdfghjk \\); hence it suffices to consider \\( asdfghjk>0 \\). Replacing \\( mnbvcxzl \\) by \\( 1-mnbvcxzl \\) and interchanging \\( qzxwvtnp \\) and \\( hjgrksla \\) preserves the inequality, so we may assume \\( qzxwvtnp \\geq hjgrksla \\).\n\nWe will show that for each \\( mnbvcxzl \\in(0,1) \\), the function\n\\[\npoiulkjh(asdfghjk)=qzxwvtnp \\frac{\\sinh asdfghjk mnbvcxzl}{\\sinh asdfghjk}+hjgrksla \\frac{\\sinh asdfghjk(1-mnbvcxzl)}{\\sinh asdfghjk}-qzxwvtnp^{mnbvcxzl} hjgrksla^{1-mnbvcxzl}\n\\]\nof \\( asdfghjk \\) is decreasing for \\( asdfghjk>0 \\). For this, it suffices to show that for each \\( \\plmoknij \\in(0,1) \\), the function\n\\[\nqazxswed(asdfghjk)=\\frac{\\sinh \\plmoknij asdfghjk}{\\sinh asdfghjk}\n\\]\nis decreasing for \\( asdfghjk>0 \\), since \\( poiulkjh(asdfghjk) \\) is a constant plus a sum of positive multiples of two such functions \\( qazxswed \\). Differentiating with respect to \\( asdfghjk \\) yields\n\\[\nqazxswed^{\\prime}(asdfghjk)=\\frac{\\plmoknij \\cosh (\\plmoknij asdfghjk) \\sinh asdfghjk-\\sinh (\\plmoknij asdfghjk) \\cosh asdfghjk}{\\sinh ^{2} asdfghjk}\n\\]\nwhich for \\( asdfghjk>0 \\) is negative if and only if the value of the function\n\\[\nrfvbgtky(asdfghjk)=\\plmoknij \\tanh asdfghjk-\\tanh (\\plmoknij asdfghjk)\n\\]\nis negative. The negativity of \\( rfvbgtky(asdfghjk) \\) for \\( asdfghjk>0 \\) follows from \\( rfvbgtky(0)=0 \\) and the negativity of\n\\[\nrfvbgtky^{\\prime}(asdfghjk)=\\frac{\\plmoknij}{\\cosh ^{2} asdfghjk}-\\frac{\\plmoknij}{\\cosh ^{2} \\plmoknij asdfghjk}\n\\]\nwhich in turn follows since cosh is increasing on \\( \\mathbb{R}^{+} \\). Thus \\( poiulkjh(asdfghjk) \\) is decreasing for \\( asdfghjk>0 \\).\n\nIf \\( qzxwvtnp>hjgrksla \\) then \\( poiulkjh(asdfghjk) \\) is zero at \\( asdfghjk=\\ln (qzxwvtnp / hjgrksla) \\). If \\( qzxwvtnp=hjgrksla \\) then \\( \\lim _{asdfghjk \\rightarrow 0^{+}} poiulkjh(asdfghjk)=0 \\) by L'H\\'ospital's Rule [Spv, Ch. 11, Theorem 9]. Since \\( poiulkjh(asdfghjk) \\) is decreasing for \\( asdfghjk>0 \\), in both cases we have \\( poiulkjh(asdfghjk) \\geq 0 \\) for \\( 0<asdfghjk \\leq \\ln (qzxwvtnp / hjgrksla) \\) and \\( poiulkjh(asdfghjk)<0 \\) for \\( asdfghjk>\\ln (qzxwvtnp / hjgrksla) \\), as desired.\n\nSolution 2. As before, we may assume \\( asdfghjk>0 \\) and \\( qzxwvtnp \\geq hjgrksla \\). Taking \\( \\ln \\) of each side and rearranging, we find that the inequality is equivalent to\n\\[\nmnbvcxzl(\\ln (qzxwvtnp \\sinh asdfghjk))+(1-mnbvcxzl)(\\ln (hjgrksla \\sinh asdfghjk)) \\leq \\ln (qzxwvtnp \\sinh asdfghjk mnbvcxzl+hjgrksla \\sinh asdfghjk(1-mnbvcxzl)),\n\\]\nwhich, if we define\n\\[\nqazxswed(mnbvcxzl)=qzxwvtnp \\sinh asdfghjk mnbvcxzl+hjgrksla \\sinh asdfghjk(1-mnbvcxzl),\n\\]\ncan be written as\n\\[\nmnbvcxzl \\ln qazxswed(1)+(1-mnbvcxzl) \\ln qazxswed(0) \\leq \\ln qazxswed(mnbvcxzl)\n\\]\n\nThis will hold if \\( asdfghjk \\) is such that \\( \\ln qazxswed(mnbvcxzl) \\) is concave for \\( mnbvcxzl \\in(0,1) \\), and will fail if \\( \\ln qazxswed(mnbvcxzl) \\) is strictly convex on \\( (0,1) \\).\n\nWe compute\n\\[\n\\begin{aligned}\n\\frac{d}{d mnbvcxzl} \\ln qazxswed= & \\frac{qazxswed^{\\prime}}{qazxswed} \\\\\n\\frac{d^{2}}{d mnbvcxzl^{2}} \\ln qazxswed= & \\frac{qazxswed qazxswed^{\\prime \\prime}-\\left(qazxswed^{\\prime}\\right)^{2}}{qazxswed^{2}} \\\\\nqazxswed^{\\prime}= & qzxwvtnp asdfghjk \\cosh asdfghjk mnbvcxzl-hjgrksla asdfghjk \\cosh asdfghjk(1-mnbvcxzl) \\\\\nqazxswed^{\\prime \\prime}= & qzxwvtnp asdfghjk^{2} \\sinh asdfghjk mnbvcxzl+hjgrksla asdfghjk^{2} \\sinh asdfghjk(1-mnbvcxzl) \\\\\nqazxswed qazxswed^{\\prime \\prime}-\\left(qazxswed^{\\prime}\\right)^{2}= & \\left(qzxwvtnp^{2}+hjgrksla^{2}\\right) asdfghjk^{2}\\left(\\sinh ^{2} asdfghjk mnbvcxzl-\\cosh ^{2} asdfghjk mnbvcxzl\\right) \\\\\n& +2 qzxwvtnp hjgrksla asdfghjk^{2}(\\sinh asdfghjk mnbvcxzl \\sinh asdfghjk(1-mnbvcxzl)+\\cosh asdfghjk mnbvcxzl \\cosh asdfghjk(1-mnbvcxzl)) \\\\\n= & \\left(qzxwvtnp^{2}+hjgrksla^{2}\\right) asdfghjk^{2}(-1)+2 qzxwvtnp hjgrksla asdfghjk^{2} \\cosh (asdfghjk mnbvcxzl+asdfghjk(1-mnbvcxzl)) \\\\\n= & asdfghjk^{2}\\left(-qzxwvtnp^{2}-hjgrksla^{2}+2 qzxwvtnp hjgrksla \\cosh asdfghjk\\right)\n\\end{aligned}\n\\]\no \\( \\frac{d^{2}}{d mnbvcxzl^{2}} \\ln qazxswed \\) has constant sign for \\( mnbvcxzl \\in(0,1) \\), and the following are equivalent:\n\\[\n\\begin{aligned}\n\\frac{d^{2}}{d mnbvcxzl^{2}} \\ln qazxswed & \\leq 0 \\quad \\text { for } mnbvcxzl \\in(0,1) \\\\\n-qzxwvtnp^{2}-hjgrksla^{2}+2 qzxwvtnp hjgrksla \\cosh asdfghjk & \\leq 0 \\\\\n-qzxwvtnp^{2}-hjgrksla^{2}+qzxwvtnp hjgrksla e^{asdfghjk}+qzxwvtnp hjgrksla e^{-asdfghjk} & \\leq 0 \\\\\n-\\left(qzxwvtnp-e^{asdfghjk} hjgrksla\\right)\\left(qzxwvtnp-e^{-asdfghjk} hjgrksla\\right) & \\leq 0 \\\\\n-\\left(qzxwvtnp-e^{asdfghjk} hjgrksla\\right) & \\leq 0 \\quad(\\text { since } asdfghjk>0 \\text { and } qzxwvtnp \\geq hjgrksla>0) \\\\\nasdfghjk & \\leq \\ln (qzxwvtnp / hjgrksla) .\n\\end{aligned}\n\\]\n"
    },
    "kernel_variant": {
      "question": "Let p and q be positive real numbers. Determine the largest non–negative real number d (expressed in terms of p and q) such that for all real numbers v and y with\n\na)   0 < |v| ≤ d,  and\nb)   0 ≤ y ≤ 1,\n\nthe inequality\n\n      p^{y}\n      q^{1-y}\n      ≤ p·\\frac{\\sinh (v y)}{\\sinh v}\n        + q·\\frac{\\sinh (v(1-y))}{\\sinh v}\n\nholds.  (Throughout, \\(\\sinh t=(e^{t}-e^{-t})/2\\).)",
      "solution": "We prove that\nd = |\\ln(p/q)|\nis the required extremal value.\n\nStep 1  (Symmetry).\nBecause sinh is odd, the right–hand side is an even function of v, so it suffices to treat v>0.  Interchanging p and q while replacing y by 1−y leaves the inequality unchanged; hence we may assume p≥q and finally restore symmetry by taking an absolute value.\n\nStep 2  (Auxiliary function).\nFix y∈[0,1] and define\n  F(v)=p·\\frac{\\sinh(vy)}{\\sinh v}+q·\\frac{\\sinh(v(1-y))}{\\sinh v}−p^{y}q^{1-y}, v>0.\nOur goal is to determine the sign of F.\n\nStep 3  (Monotonicity of a basic quotient).\nFor 0<α<1 put f(v)=\\sinh(αv)/\\sinh v.  Differentiation gives\n  f'(v)=\\frac{α\\cosh(αv)\\sinh v−\\sinh(αv)\\cosh v}{\\sinh^{2}v}\n       =\\frac{\\cosh(αv)\\cosh v}{\\sinh^{2}v}\\,[α\\tanh v−\\tanh(αv)].\nThe factor (cosh(αv)cosh v)/sinh²v is positive for v>0, so the sign of f' equals the sign of g(v)=α tanh v−tanh(αv).  Because tanh is increasing with tanh0=0 we have g(v)<0 for every v>0; hence f'(v)<0, i.e. f is strictly decreasing.  Consequently, for any p,q>0 and 0<α<1, the linear combination p f(v)+q f(1−α)(v) is also strictly decreasing.\n\nStep 4  (Monotonicity of F).\nBoth fractions occurring in F are of the type f, so F'(v)<0 for every v>0; thus F is strictly decreasing on (0,∞).\n\nStep 5  (The unique zero of F).\nBecause p≥q>0 we can write p=q e^{v₀} with v₀=\\ln(p/q)≥0.  Substituting v=v₀ we get\n  p·\\frac{\\sinh(v₀y)}{\\sinh v₀}+q·\\frac{\\sinh(v₀(1−y))}{\\sinh v₀}=p^{y}q^{1−y},\nso F(v₀)=0.  (When p=q we have v₀=0 and lim_{v→0⁺}F(v)=0.)\n\nStep 6  (Positivity range and maximal d).\nSince F is strictly decreasing and F(v₀)=0,\n  F(v)≥0 for 0<v≤v₀, and F(v)<0 for v>v₀.\nHence the stated inequality holds exactly for 0<|v|≤v₀ when p≥q.  Restoring symmetry by swapping p and q yields the largest admissible number\n  d=|v₀|=|\\ln(p/q)|.\n\nSpecial case p=q.  Then v₀=0, so the largest d is 0; this is allowed because for d=0 the hypothesis 0<|v|≤d is never satisfied and the implication is vacuously true.\n\nTherefore the maximal value is\nd = |\\ln(p/q)|.",
      "_meta": {
        "core_steps": [
          "Exploit symmetry: since the RHS is even in u and invariant under (a,b,x)↔(b,a,1−x), assume u>0 and a≥b.",
          "Introduce F(u)=a·sinh(ux)/sinh u + b·sinh(u(1−x))/sinh u − a^{x}b^{1−x}.",
          "Prove F′(u)<0 for u>0 by showing f(u)=sinh(αu)/sinh u (0<α<1) is decreasing; this follows from derivative analysis with tanh and the monotonicity of cosh.",
          "Note F(ln(a/b))=0 (and F→0 as u→0⁺); with F decreasing, we get F(u)≥0 for 0<u≤ln(a/b) and F(u)<0 for u>ln(a/b).",
          "Hence the inequality holds exactly when 0<|u|≤|ln(a/b)|, so the maximal constant is c=|ln(a/b)|."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Because the RHS is an even function, the absolute–value sign in the range for u is optional; one may restrict to u≥0 instead.",
            "original": "|u| in the condition 0<|u|≤c"
          },
          "slot2": {
            "description": "The open interval for x can be replaced by the closed interval; the endpoints x=0,1 give equality and do not affect the argument.",
            "original": "0 < x < 1"
          },
          "slot3": {
            "description": "‘sinh’ could be replaced by any odd, increasing function h with the property that h(αu)/h(u) decreases in u for 0<α<1; the derivative-based monotonicity proof is unchanged.",
            "original": "sinh"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}