{
  "index": "1975-B-3",
  "type": "ALG",
  "tag": [
    "ALG",
    "COMB"
  ],
  "difficulty": "",
  "question": "B-3. Let \\( s_{k}\\left(a_{1}, \\ldots, a_{n}\\right) \\) denote the \\( k \\)-th elementary symmetric function of \\( a_{1}, \\ldots, a_{n} \\). With \\( k \\) held fixed, find the supremum (or least upper bound) \\( M_{k} \\) of\n\\[\ns_{k}\\left(a_{1}, \\ldots, a_{n}\\right) /\\left[s_{1}\\left(a_{1}, \\ldots, a_{n}\\right)\\right]^{k}\n\\]\nfor arbitrary \\( n \\geqq k \\) and arbitrary \\( n \\)-tuples \\( a_{1}, \\ldots, a_{n} \\) of positive real numbers.\n[The symmetric function \\( s_{k}\\left(a_{1}, \\ldots, a_{n}\\right) \\) is the sum of all \\( k \\)-fold products of the variables \\( a_{1}, \\ldots, a_{n} \\). Thus, for example:\n\\[\n\\begin{array}{c}\ns_{1}\\left(a_{1}, \\ldots, a_{n}\\right)=a_{1}+a_{2}+\\ldots+a_{n} \\\\\ns_{3}\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)=a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{1} a_{3} a_{4}+a_{2} a_{3} a_{4}\n\\end{array}\n\\]\n\nIt should be remarked that the supremum \\( \\boldsymbol{M}_{\\boldsymbol{k}} \\) is never attained; it is approached arbitrarily closely when, for fixed \\( k \\), the number \\( n \\) of variables increases without bound, and the values \\( a_{i}>0 \\) are suitably chosen.]",
  "solution": "B-3.\nIn the expansion of \\( s_{1}^{k}=\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{k} \\), every term of \\( s_{k} \\) appears with \\( k \\) ! as coefficient and the other coefficients are nonnegative. Hence \\( s_{k} / s_{1}^{k} \\leqq 1 / k \\) !\n\nIf we let each \\( a_{i}=1 \\),\n\\[\n\\frac{s_{k}}{s_{1}^{k}}=\\binom{n}{k} / n^{k}=\\frac{n(n-1) \\cdots(n-k+1)}{k!n^{k}}=\\frac{1}{k!}\\left(1-\\frac{1}{n}\\right)\\left(1-\\frac{2}{n}\\right) \\cdots\\left(1-\\frac{k-1}{n}\\right),\n\\]\nwhich approaches \\( 1 / k \\) ! as \\( k \\) is held fixed and \\( n \\) goes to infinity. These facts show that the supremum \\( M_{k} \\) is \\( 1 / k! \\).",
  "vars": [
    "a_1",
    "a_2",
    "a_3",
    "a_4",
    "a_i",
    "a_n",
    "n"
  ],
  "params": [
    "k",
    "s_k",
    "s_1",
    "M_k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "a_1": "alphaone",
        "a_2": "alphatwo",
        "a_3": "alphathree",
        "a_4": "alphafour",
        "a_i": "alphagen",
        "a_n": "alphanlast",
        "n": "varcount",
        "k": "combindex",
        "s_k": "symmkth",
        "s_1": "symmones",
        "M_k": "supremkth"
      },
      "question": "B-3. Let \\( symmkth\\left(alphaone, \\ldots, alphanlast\\right) \\) denote the \\( combindex \\)-th elementary symmetric function of \\( alphaone, \\ldots, alphanlast \\). With \\( combindex \\) held fixed, find the supremum (or least upper bound) \\( supremkth \\) of\n\\[\nsymmkth\\left(alphaone, \\ldots, alphanlast\\right) /\\left[ symmones\\left(alphaone, \\ldots, alphanlast\\right) \\right]^{combindex}\n\\]\nfor arbitrary \\( varcount \\geqq combindex \\) and arbitrary varcount-tuples \\( alphaone, \\ldots, alphanlast \\) of positive real numbers.\n[The symmetric function \\( symmkth\\left(alphaone, \\ldots, alphanlast\\right) \\) is the sum of all \\( combindex \\)-fold products of the variables \\( alphaone, \\ldots, alphanlast \\). Thus, for example:\n\\[\n\\begin{array}{c}\nsymmones\\left(alphaone, \\ldots, alphanlast\\right)=alphaone+alphatwo+\\ldots+alphanlast \\\\\ns_{3}\\left(alphaone, alphatwo, alphathree, alphafour\\right)=alphaone\\,alphatwo\\,alphathree+alphaone\\,alphatwo\\,alphafour+alphaone\\,alphathree\\,alphafour+alphatwo\\,alphathree\\,alphafour\n\\end{array}\n\\]\n\nIt should be remarked that the supremum \\( \\boldsymbol{supremkth} \\) is never attained; it is approached arbitrarily closely when, for fixed \\( combindex \\), the number \\( varcount \\) of variables increases without bound, and the values \\( alphagen>0 \\) are suitably chosen.]",
      "solution": "B-3.\nIn the expansion of \\( symmones^{combindex}=\\left(alphaone+alphatwo+\\cdots+alphanlast\\right)^{combindex} \\), every term of \\( symmkth \\) appears with \\( combindex! \\) as coefficient and the other coefficients are nonnegative. Hence \\( symmkth / symmones^{combindex} \\leqq 1 / combindex! \\).\n\nIf we let each \\( alphagen=1 \\),\n\\[\n\\frac{symmkth}{symmones^{combindex}}=\\binom{varcount}{combindex} / varcount^{combindex}=\\frac{varcount(varcount-1) \\cdots (varcount-combindex+1)}{combindex!\\,varcount^{combindex}}=\\frac{1}{combindex!}\\left(1-\\frac{1}{varcount}\\right)\\left(1-\\frac{2}{varcount}\\right) \\cdots \\left(1-\\frac{combindex-1}{varcount}\\right),\n\\]\nwhich approaches \\( 1 / combindex! \\) as \\( combindex \\) is held fixed and \\( varcount \\) goes to infinity. These facts show that the supremum \\( supremkth \\) is \\( 1 / combindex! \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "a_1": "arrowroot",
        "a_2": "buttercup",
        "a_3": "buttermilk",
        "a_4": "chandelier",
        "a_i": "dragonfly",
        "a_n": "scarecrow",
        "n": "horseshoe",
        "k": "windchime",
        "s_k": "grassland",
        "s_1": "riverbank",
        "M_k": "cornstalk"
      },
      "question": "B-3. Let \\( grassland\\left(arrowroot, \\ldots, scarecrow\\right) \\) denote the \\( windchime \\)-th elementary symmetric function of \\( arrowroot, \\ldots, scarecrow \\). With \\( windchime \\) held fixed, find the supremum (or least upper bound) \\( cornstalk \\) of\n\\[\ngrassland\\left(arrowroot, \\ldots, scarecrow\\right) /\\left[riverbank\\left(arrowroot, \\ldots, scarecrow\\right)\\right]^{windchime}\n\\]\nfor arbitrary \\( horseshoe \\geqq windchime \\) and arbitrary \\( horseshoe \\)-tuples \\( arrowroot, \\ldots, scarecrow \\) of positive real numbers.\n[The symmetric function \\( grassland\\left(arrowroot, \\ldots, scarecrow\\right) \\) is the sum of all \\( windchime \\)-fold products of the variables \\( arrowroot, \\ldots, scarecrow \\). Thus, for example:\n\\[\n\\begin{array}{c}\nriverbank\\left(arrowroot, \\ldots, scarecrow\\right)=arrowroot+buttercup+\\ldots+scarecrow \\\\\ns_{3}\\left(arrowroot, buttercup, buttermilk, chandelier\\right)=arrowroot buttercup buttermilk+arrowroot buttercup chandelier+arrowroot buttermilk chandelier+buttercup buttermilk chandelier\n\\end{array}\n\\]\n\nIt should be remarked that the supremum \\( \\boldsymbol{cornstalk} \\) is never attained; it is approached arbitrarily closely when, for fixed \\( windchime \\), the number \\( horseshoe \\) of variables increases without bound, and the values \\( dragonfly>0 \\) are suitably chosen.]",
      "solution": "B-3.\nIn the expansion of \\( riverbank^{windchime}=\\left(arrowroot+buttercup+\\cdots+scarecrow\\right)^{windchime} \\), every term of \\( grassland \\) appears with \\( windchime \\) ! as coefficient and the other coefficients are nonnegative. Hence \\( grassland / riverbank^{windchime} \\leqq 1 / windchime \\) !\n\nIf we let each \\( dragonfly=1 \\),\n\\[\n\\frac{grassland}{riverbank^{windchime}}=\\binom{horseshoe}{windchime} / horseshoe^{windchime}=\\frac{horseshoe(horseshoe-1) \\cdots(horseshoe-windchime+1)}{windchime!horseshoe^{windchime}}=\\frac{1}{windchime!}\\left(1-\\frac{1}{horseshoe}\\right)\\left(1-\\frac{2}{horseshoe}\\right) \\cdots\\left(1-\\frac{windchime-1}{horseshoe}\\right),\n\\]\nwhich approaches \\( 1 / windchime! \\) as \\( windchime \\) is held fixed and \\( horseshoe \\) goes to infinity. These facts show that the supremum \\( cornstalk \\) is \\( 1 / windchime! \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "a_1": "negvalueone",
        "a_2": "negvaluetwo",
        "a_3": "negvaluethree",
        "a_4": "negvaluefour",
        "a_i": "negvaluei",
        "a_n": "negvaluen",
        "n": "scarcitycount",
        "k": "fluidindex",
        "s_k": "antisymfun",
        "s_1": "antisymfirst",
        "M_k": "infimumvalue"
      },
      "question": "B-3. Let \\( antisymfun\\left(negvalueone, \\ldots, negvaluen\\right) \\) denote the \\( fluidindex \\)-th elementary symmetric function of \\( negvalueone, \\ldots, negvaluen \\). With \\( fluidindex \\) held fixed, find the supremum (or least upper bound) \\( infimumvalue \\) of\n\\[\nantisymfun\\left(negvalueone, \\ldots, negvaluen\\right) /\\left[antisymfirst\\left(negvalueone, \\ldots, negvaluen\\right)\\right]^{fluidindex}\n\\]\nfor arbitrary \\( scarcitycount \\geqq fluidindex \\) and arbitrary \\( scarcitycount \\)-tuples \\( negvalueone, \\ldots, negvaluen \\) of positive real numbers.\n[The symmetric function \\( antisymfun\\left(negvalueone, \\ldots, negvaluen\\right) \\) is the sum of all \\( fluidindex \\)-fold products of the variables \\( negvalueone, \\ldots, negvaluen \\). Thus, for example:\n\\[\n\\begin{array}{c}\nantisymfirst\\left(negvalueone, \\ldots, negvaluen\\right)=negvalueone+negvaluetwo+\\ldots+negvaluen \\\\\ns_{3}\\left(negvalueone, negvaluetwo, negvaluethree, negvaluefour\\right)=negvalueone negvaluetwo negvaluethree+negvalueone negvaluetwo negvaluefour+negvalueone negvaluethree negvaluefour+negvaluetwo negvaluethree negvaluefour\n\\end{array}\n\\]\n\nIt should be remarked that the supremum \\( \\boldsymbol{infimumvalue} \\) is never attained; it is approached arbitrarily closely when, for fixed \\( fluidindex \\), the number \\( scarcitycount \\) of variables increases without bound, and the values \\( negvaluei>0 \\) are suitably chosen.]",
      "solution": "B-3.\nIn the expansion of \\( antisymfirst^{fluidindex}=\\left(negvalueone+negvaluetwo+\\cdots+negvaluen\\right)^{fluidindex} \\), every term of \\( antisymfun \\) appears with \\( fluidindex \\) ! as coefficient and the other coefficients are nonnegative. Hence \\( antisymfun / antisymfirst^{fluidindex} \\leqq 1 / fluidindex \\) !\n\nIf we let each \\( negvaluei=1 \\),\n\\[\n\\frac{antisymfun}{antisymfirst^{fluidindex}}=\\binom{scarcitycount}{fluidindex} / scarcitycount^{fluidindex}=\\frac{scarcitycount(scarcitycount-1) \\cdots(scarcitycount-fluidindex+1)}{fluidindex!scarcitycount^{fluidindex}}=\\frac{1}{fluidindex!}\\left(1-\\frac{1}{scarcitycount}\\right)\\left(1-\\frac{2}{scarcitycount}\\right) \\cdots\\left(1-\\frac{fluidindex-1}{scarcitycount}\\right),\n\\]\nwhich approaches \\( 1 / fluidindex \\) ! as \\( fluidindex \\) is held fixed and \\( scarcitycount \\) goes to infinity. These facts show that the supremum \\( infimumvalue \\) is \\( 1 / fluidindex! \\)."
    },
    "garbled_string": {
      "map": {
        "a_1": "qzxwvtnp",
        "a_2": "hjgrksla",
        "a_3": "mndyfeop",
        "a_4": "vklisuez",
        "a_i": "ptawxrob",
        "a_n": "clymbgfa",
        "n": "zwekfuni",
        "k": "srdvmxqe",
        "s_k": "oyqzhjrn",
        "s_1": "gplcxadu",
        "M_k": "xytbrmne"
      },
      "question": "B-3. Let \\( oyqzhjrn\\left(qzxwvtnp, \\ldots, clymbgfa\\right) \\) denote the \\( srdvmxqe \\)-th elementary symmetric function of \\( qzxwvtnp, \\ldots, clymbgfa \\). With \\( srdvmxqe \\) held fixed, find the supremum (or least upper bound) \\( xytbrmne \\) of\n\\[\noyqzhjrn\\left(qzxwvtnp, \\ldots, clymbgfa\\right) /\\left[gplcxadu\\left(qzxwvtnp, \\ldots, clymbgfa\\right)\\right]^{srdvmxqe}\n\\]\nfor arbitrary \\( zwekfuni \\geqq srdvmxqe \\) and arbitrary \\( zwekfuni \\)-tuples \\( qzxwvtnp, \\ldots, clymbgfa \\) of positive real numbers.\n[The symmetric function \\( oyqzhjrn\\left(qzxwvtnp, \\ldots, clymbgfa\\right) \\) is the sum of all \\( srdvmxqe \\)-fold products of the variables \\( qzxwvtnp, \\ldots, clymbgfa \\). Thus, for example:\n\\[\n\\begin{array}{c}\ngplcxadu\\left(qzxwvtnp, \\ldots, clymbgfa\\right)=qzxwvtnp+hjgrksla+\\ldots+clymbgfa \\\\\ns_{3}\\left(qzxwvtnp, hjgrksla, mndyfeop, vklisuez\\right)=qzxwvtnp hjgrksla mndyfeop+qzxwvtnp hjgrksla vklisuez+qzxwvtnp mndyfeop vklisuez+hjgrksla mndyfeop vklisuez\n\\end{array}\n\\]\n\nIt should be remarked that the supremum \\( \\boldsymbol{xytbrmne} \\) is never attained; it is approached arbitrarily closely when, for fixed \\( srdvmxqe \\), the number \\( zwekfuni \\) of variables increases without bound, and the values \\( ptawxrob>0 \\) are suitably chosen.]",
      "solution": "B-3.\nIn the expansion of \\( gplcxadu^{srdvmxqe}=\\left(qzxwvtnp+hjgrksla+\\cdots+clymbgfa\\right)^{srdvmxqe} \\), every term of \\( oyqzhjrn \\) appears with \\( srdvmxqe \\) ! as coefficient and the other coefficients are nonnegative. Hence \\( oyqzhjrn / gplcxadu^{srdvmxqe} \\leqq 1 / srdvmxqe \\) !\n\nIf we let each \\( ptawxrob=1 \\),\n\\[\n\\frac{oyqzhjrn}{gplcxadu^{srdvmxqe}}=\\binom{zwekfuni}{srdvmxqe} / zwekfuni^{srdvmxqe}=\\frac{zwekfuni(zwekfuni-1) \\cdots(zwekfuni-srdvmxqe+1)}{srdvmxqe!zwekfuni^{srdvmxqe}}=\\frac{1}{srdvmxqe!}\\left(1-\\frac{1}{zwekfuni}\\right)\\left(1-\\frac{2}{zwekfuni}\\right) \\cdots\\left(1-\\frac{srdvmxqe-1}{zwekfuni}\\right),\n\\]\nwhich approaches \\( 1 / srdvmxqe \\) ! as \\( srdvmxqe \\) is held fixed and \\( zwekfuni \\) goes to infinity. These facts show that the supremum \\( xytbrmne \\) is \\( 1 / srdvmxqe! \\)."
    },
    "kernel_variant": {
      "question": "Let $m\\ge 2$ be an integer and let  \n\n\\[\n1\\le k_{1}\\le k_{2}\\le\\dots\\le k_{m}\n\\]\n\nbe (not necessarily distinct) positive integers.  \nLet $\\theta_{1},\\dots ,\\theta_{m}$ be positive rational numbers and put  \n\n\\[\n\\Theta=\\sum_{j=1}^{m}\\theta_{j}k_{j}.\n\\]\n\nFor every integer $n\\ge k_{m}$ and every $n$–tuple  \n\n\\[\n(q_{1},\\dots ,q_{n})\\in\\mathbb Q_{\\ge 0}^{\\,n}\\setminus\\{(0,\\dots ,0)\\},\n\\]\n\ndenote by $E_{j}(q_{1},\\dots ,q_{n})$ the $j$-th elementary symmetric function of the variables $q_{1},\\dots ,q_{n}$.  Define  \n\n\\[\nR(q_{1},\\dots ,q_{n})=\n\\frac{E_{k_{1}}(q_{1},\\dots ,q_{n})^{\\theta_{1}}\n      \\,\\cdots\\,\n      E_{k_{m}}(q_{1},\\dots ,q_{n})^{\\theta_{m}}}\n     {E_{1}(q_{1},\\dots ,q_{n})^{\\;\\Theta}}.\n\\]\n\nDetermine the exact value  \n\n\\[\nM(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n=\\sup_{\\,n\\ge k_{m}}\n \\;\\sup_{\\,q\\neq 0} R(q)\n\\]\n\nand decide precisely for which parameter sets $(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})$ the supremum is actually attained, that is, there exist a finite $n$ and a non–zero $n$-tuple $q$ with  \n$R(q)=M(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})$.",
      "solution": "Throughout we abbreviate $E_{j}(q_{1},\\dots ,q_{n})$ by $E_{j}$ and write  \n$q=(q_{1},\\dots ,q_{n})$.\n\n\\textbf{Step 1.  A universal upper bound}  \n\nFor every integer $k\\ge 1$ the well–known identity  \n\\[\nE_{1}^{\\,k}=k!\\,E_{k}+S_{k},\n\\]\nwhere $S_{k}$ is a polynomial whose monomials all contain repeated\nindices, implies $S_{k}(q)\\ge 0$ whenever $q_{i}\\ge 0$.  Hence  \n\\[\nE_{k}\\le\\frac{E_{1}^{\\,k}}{k!},\n\\qquad\\text{and}\\qquad\nE_{k}<\\frac{E_{1}^{\\,k}}{k!}\\quad\\text{if }k\\ge 2\\text{ and }q\\neq 0.\n\\tag{1}\n\\]\n\nWrite each $\\theta_{j}$ in lowest terms as $r_{j}/s$\n($r_{j}\\in\\mathbb N^{+}$, $s\\in\\mathbb N^{+}$) and set $r=\\sum_{j=1}^{m}r_{j}$.\nRaising $R(q)$ to the $s$-th power clears the rational exponents:\n\\[\nR(q)^{s}\n=\\frac{\\prod_{j=1}^{m}E_{k_{j}}^{\\,r_{j}}}{E_{1}^{\\,s\\Theta}}.\n\\tag{2}\n\\]\nApplying inequality (1) to each factor and using  \n$s\\Theta=\\sum_{j=1}^{m}r_{j}k_{j}$ gives\n\\[\nE_{k_{j}}^{\\,r_{j}}\\le\n\\frac{E_{1}^{\\,r_{j}k_{j}}}{(k_{j}!)^{r_{j}}},\n\\qquad j=1,\\dots ,m.\n\\tag{3}\n\\]\nMultiplying the $m$ inequalities (3) and comparing with (2) yields\n\\[\nR(q)^{s}\\le\n\\Bigl(\\prod_{j=1}^{m}(k_{j}!)^{-r_{j}}\\Bigr),\n\\]\nhence\n\\[\nR(q)\\le\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}.\n\\tag{4}\n\\]\n\n\\textbf{Step 2.  Sharpness of the bound}  \n\nFix $n\\ge k_{m}$ and choose $q_{1}=\\dots =q_{n}=1$.  Then\n$E_{k_{j}}=\\binom{n}{k_{j}}$, so\n\\[\nR_{n}:=R(1,\\dots ,1)\n     =\\frac{\\prod_{j=1}^{m}\\binom{n}{k_{j}}^{\\theta_{j}}}{n^{\\Theta}}.\n\\tag{5}\n\\]\nBy Stirling’s formula\n$\\binom{n}{k_{j}}=\\dfrac{n^{k_{j}}}{k_{j}!}\\bigl(1+O(1/n)\\bigr)$,\nand substitution into (5) gives\n\\[\nR_{n}=\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}\n       \\bigl(1+O(1/n)\\bigr).\n\\]\nLetting $n\\to\\infty$ we obtain\n\\[\n\\lim_{n\\to\\infty}R_{n}\n      =\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}},\n\\tag{6}\n\\]\nwhich coincides with the upper bound (4).  Therefore  \n\n\\[\nM(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n=\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}.\n\\tag{7}\n\\]\n\n\\textbf{Step 3.  When is the supremum attained?}\n\n\\emph{(i)  If at least one $k_{j}\\ge 2$.}  \nChoose such an index $j_{0}$.  \nThe strict version of (1) for $k=k_{j_{0}}$ implies\n\\[\nE_{k_{j_{0}}}<\\frac{E_{1}^{\\,k_{j_{0}}}}{k_{j_{0}}!}\n\\quad\\text{for every non–zero }q.\n\\]\nKeeping equality for the remaining indices and repeating the argument\nthat led to (4) we now get\n\\[\nR(q)<\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}\n      =M(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n\\quad\\text{for every finite }n\\text{ and }q\\neq 0.\n\\]\nThus the supremum in (7) is not attained.\n\n\\emph{(ii)  If $k_{1}=k_{2}=\\dots =k_{m}=1$.}  \nThen $\\Theta=\\sum_{j=1}^{m}\\theta_{j}$ and $E_{1}$ appears in both\nnumerator and denominator with the same exponent, so $R(q)=1$ for every\n$n$ and every non–zero $q$.  Consequently\n\\[\nM(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})=1,\n\\]\nand the supremum is attained for \\emph{every} admissible $n$ and\n$q$.\n\n\\textbf{Step 4.  Final characterisation}  \n\nThe above discussion shows that\n\n\\[\n\\text{The supremum (7) is attained\n      if and only if }k_{1}=k_{2}=\\dots =k_{m}=1.\n\\]\n\nIn all other parameter configurations (equivalently, whenever\nat least one $k_{j}\\ge 2$) the value in (7) is approached only\nin the limit $n\\to\\infty$ and is never reached by any finite\nchoice of $n$ and $q$.\n\n\\[\n\\boxed{\n\\;M(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n  =\\displaystyle\\prod_{j=1}^{m}\\Bigl(\\tfrac{1}{k_{j}!}\\Bigr)^{\\theta_{j}},\n\\;\n\\text{and}\\;\nR(q)=M\\text{ is attainable }\n\\Longleftrightarrow k_{1}=\\dots =k_{m}=1\n\\;}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.621031",
        "was_fixed": false,
        "difficulty_analysis": "• Multiple interacting objects:  the problem now involves m different\nelementary symmetric functions raised to independent rational powers,\ninstead of a single one.\n\n• Number-theoretic complication:  rational exponents force a\nclearing-denominator argument and introduce an additional parameter s,\nso the proof must cope with non-integral powers.\n\n• Combinatorial explosion:  bounding the product\nE_{k₁}^{r₁} … E_{kₘ}^{rₘ}\nrequires tracking the coefficients of a huge family of monomials and\nexplaining why they are all simultaneously dominated by the expansion of\nE₁^{Θs}.  This calls for a refined combinatorial–algebraic argument,\nwell beyond the single counting step of the original problem.\n\n• Achievability analysis:  deciding when equality can actually occur\nrequires a subtle inspection of the binomial-coefficient factor\nC(n,k_j)/n^{k_j} < 1/k_j! and a case distinction according to whether\nsome k_j > 1, adding an extra layer of reasoning absent from the\nprototype.\n\nCollectively these elements make the enhanced variant appreciably\nharder; it demands mastery of multinomial expansions, asymptotic\nestimates, clearing denominators for rational powers, and a careful\nextremal analysis, rather than the single elementary inequality that\nsolves the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $m\\ge 2$ be an integer and let  \n\n\\[\n1\\le k_{1}\\le k_{2}\\le\\dots\\le k_{m}\n\\]\n\nbe (not necessarily distinct) positive integers.  \nLet $\\theta_{1},\\dots ,\\theta_{m}$ be positive rational numbers and put  \n\n\\[\n\\Theta=\\sum_{j=1}^{m}\\theta_{j}k_{j}.\n\\]\n\nFor every integer $n\\ge k_{m}$ and every $n$–tuple  \n\n\\[\n(q_{1},\\dots ,q_{n})\\in\\mathbb Q_{\\ge 0}^{\\,n}\\setminus\\{(0,\\dots ,0)\\},\n\\]\n\ndenote by $E_{j}(q_{1},\\dots ,q_{n})$ the $j$-th elementary symmetric function of the variables $q_{1},\\dots ,q_{n}$.  Define  \n\n\\[\nR(q_{1},\\dots ,q_{n})=\n\\frac{E_{k_{1}}(q_{1},\\dots ,q_{n})^{\\theta_{1}}\n      \\,\\cdots\\,\n      E_{k_{m}}(q_{1},\\dots ,q_{n})^{\\theta_{m}}}\n     {E_{1}(q_{1},\\dots ,q_{n})^{\\;\\Theta}}.\n\\]\n\nDetermine the exact value  \n\n\\[\nM(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n=\\sup_{\\,n\\ge k_{m}}\n \\;\\sup_{\\,q\\neq 0} R(q)\n\\]\n\nand decide precisely for which parameter sets $(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})$ the supremum is actually attained, that is, there exist a finite $n$ and a non–zero $n$-tuple $q$ with  \n$R(q)=M(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})$.",
      "solution": "Throughout we abbreviate $E_{j}(q_{1},\\dots ,q_{n})$ by $E_{j}$ and write  \n$q=(q_{1},\\dots ,q_{n})$.\n\n\\textbf{Step 1.  A universal upper bound}  \n\nFor every integer $k\\ge 1$ the well–known identity  \n\\[\nE_{1}^{\\,k}=k!\\,E_{k}+S_{k},\n\\]\nwhere $S_{k}$ is a polynomial whose monomials all contain repeated\nindices, implies $S_{k}(q)\\ge 0$ whenever $q_{i}\\ge 0$.  Hence  \n\\[\nE_{k}\\le\\frac{E_{1}^{\\,k}}{k!},\n\\qquad\\text{and}\\qquad\nE_{k}<\\frac{E_{1}^{\\,k}}{k!}\\quad\\text{if }k\\ge 2\\text{ and }q\\neq 0.\n\\tag{1}\n\\]\n\nWrite each $\\theta_{j}$ in lowest terms as $r_{j}/s$\n($r_{j}\\in\\mathbb N^{+}$, $s\\in\\mathbb N^{+}$) and set $r=\\sum_{j=1}^{m}r_{j}$.\nRaising $R(q)$ to the $s$-th power clears the rational exponents:\n\\[\nR(q)^{s}\n=\\frac{\\prod_{j=1}^{m}E_{k_{j}}^{\\,r_{j}}}{E_{1}^{\\,s\\Theta}}.\n\\tag{2}\n\\]\nApplying inequality (1) to each factor and using  \n$s\\Theta=\\sum_{j=1}^{m}r_{j}k_{j}$ gives\n\\[\nE_{k_{j}}^{\\,r_{j}}\\le\n\\frac{E_{1}^{\\,r_{j}k_{j}}}{(k_{j}!)^{r_{j}}},\n\\qquad j=1,\\dots ,m.\n\\tag{3}\n\\]\nMultiplying the $m$ inequalities (3) and comparing with (2) yields\n\\[\nR(q)^{s}\\le\n\\Bigl(\\prod_{j=1}^{m}(k_{j}!)^{-r_{j}}\\Bigr),\n\\]\nhence\n\\[\nR(q)\\le\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}.\n\\tag{4}\n\\]\n\n\\textbf{Step 2.  Sharpness of the bound}  \n\nFix $n\\ge k_{m}$ and choose $q_{1}=\\dots =q_{n}=1$.  Then\n$E_{k_{j}}=\\binom{n}{k_{j}}$, so\n\\[\nR_{n}:=R(1,\\dots ,1)\n     =\\frac{\\prod_{j=1}^{m}\\binom{n}{k_{j}}^{\\theta_{j}}}{n^{\\Theta}}.\n\\tag{5}\n\\]\nBy Stirling’s formula\n$\\binom{n}{k_{j}}=\\dfrac{n^{k_{j}}}{k_{j}!}\\bigl(1+O(1/n)\\bigr)$,\nand substitution into (5) gives\n\\[\nR_{n}=\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}\n       \\bigl(1+O(1/n)\\bigr).\n\\]\nLetting $n\\to\\infty$ we obtain\n\\[\n\\lim_{n\\to\\infty}R_{n}\n      =\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}},\n\\tag{6}\n\\]\nwhich coincides with the upper bound (4).  Therefore  \n\n\\[\nM(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n=\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}.\n\\tag{7}\n\\]\n\n\\textbf{Step 3.  When is the supremum attained?}\n\n\\emph{(i)  If at least one $k_{j}\\ge 2$.}  \nChoose such an index $j_{0}$.  \nThe strict version of (1) for $k=k_{j_{0}}$ implies\n\\[\nE_{k_{j_{0}}}<\\frac{E_{1}^{\\,k_{j_{0}}}}{k_{j_{0}}!}\n\\quad\\text{for every non–zero }q.\n\\]\nKeeping equality for the remaining indices and repeating the argument\nthat led to (4) we now get\n\\[\nR(q)<\\prod_{j=1}^{m}\\Bigl(\\frac{1}{k_{j}!}\\Bigr)^{\\theta_{j}}\n      =M(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n\\quad\\text{for every finite }n\\text{ and }q\\neq 0.\n\\]\nThus the supremum in (7) is not attained.\n\n\\emph{(ii)  If $k_{1}=k_{2}=\\dots =k_{m}=1$.}  \nThen $\\Theta=\\sum_{j=1}^{m}\\theta_{j}$ and $E_{1}$ appears in both\nnumerator and denominator with the same exponent, so $R(q)=1$ for every\n$n$ and every non–zero $q$.  Consequently\n\\[\nM(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})=1,\n\\]\nand the supremum is attained for \\emph{every} admissible $n$ and\n$q$.\n\n\\textbf{Step 4.  Final characterisation}  \n\nThe above discussion shows that\n\n\\[\n\\text{The supremum (7) is attained\n      if and only if }k_{1}=k_{2}=\\dots =k_{m}=1.\n\\]\n\nIn all other parameter configurations (equivalently, whenever\nat least one $k_{j}\\ge 2$) the value in (7) is approached only\nin the limit $n\\to\\infty$ and is never reached by any finite\nchoice of $n$ and $q$.\n\n\\[\n\\boxed{\n\\;M(k_{1},\\dots ,k_{m};\\theta_{1},\\dots ,\\theta_{m})\n  =\\displaystyle\\prod_{j=1}^{m}\\Bigl(\\tfrac{1}{k_{j}!}\\Bigr)^{\\theta_{j}},\n\\;\n\\text{and}\\;\nR(q)=M\\text{ is attainable }\n\\Longleftrightarrow k_{1}=\\dots =k_{m}=1\n\\;}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.496219",
        "was_fixed": false,
        "difficulty_analysis": "• Multiple interacting objects:  the problem now involves m different\nelementary symmetric functions raised to independent rational powers,\ninstead of a single one.\n\n• Number-theoretic complication:  rational exponents force a\nclearing-denominator argument and introduce an additional parameter s,\nso the proof must cope with non-integral powers.\n\n• Combinatorial explosion:  bounding the product\nE_{k₁}^{r₁} … E_{kₘ}^{rₘ}\nrequires tracking the coefficients of a huge family of monomials and\nexplaining why they are all simultaneously dominated by the expansion of\nE₁^{Θs}.  This calls for a refined combinatorial–algebraic argument,\nwell beyond the single counting step of the original problem.\n\n• Achievability analysis:  deciding when equality can actually occur\nrequires a subtle inspection of the binomial-coefficient factor\nC(n,k_j)/n^{k_j} < 1/k_j! and a case distinction according to whether\nsome k_j > 1, adding an extra layer of reasoning absent from the\nprototype.\n\nCollectively these elements make the enhanced variant appreciably\nharder; it demands mastery of multinomial expansions, asymptotic\nestimates, clearing denominators for rational powers, and a careful\nextremal analysis, rather than the single elementary inequality that\nsolves the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}