{
  "index": "1940-B-7",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "15. Which is greater\n\\[\n(\\sqrt{n})^{\\sqrt{n+1}} \\text { or }(\\sqrt{n+1})^{\\sqrt{n}}\n\\]\nwhere \\( n>8 \\) ?",
  "solution": "Solution. \\( (\\sqrt{n})^{\\sqrt{n+1}} \\) is greater than \\( (\\sqrt{n+1})^{\\sqrt{n}} \\) for \\( n>8 \\). Consider the function \\( f(x)=(\\log x) / x \\) for \\( x>0 \\). Its derivative is \\( (1-\\log x) / x^{2} \\) which is negative for \\( x>e \\).\n\nHence, if \\( e \\leq x<y \\) we have \\( f(x)>f(y) \\), and\n\\[\nx y\\left(\\frac{\\log x}{x}\\right)>x y\\left(\\frac{\\log y}{y}\\right) .\n\\]\n\nTaking exponentials we get\n\\[\n\\boldsymbol{e}^{y \\log x}>\\boldsymbol{e}^{x \\log y},\n\\]\nthat is,\n\\[\nx^{y}>y^{x}\n\\]\nprovided \\( e \\leq x<y \\).\nIf \\( n \\geq 8 \\), then \\( e<\\sqrt{n}<\\sqrt{n+1} \\), so\n\\[\n(\\sqrt{n})^{\\sqrt{n+1}}>(\\sqrt{n+1})^{v_{n}} .\n\\]\n\nRemark. A number of interesting problems are based on the inequality (1). See for example Journal of Recreational Mathematics, vol. 2, no. 4 (October 1969), pages 255-256, where the inequality \\( e^{\\pi}>\\pi^{e} \\) and some generalizations are discussed. The inequality is related to Problem P.M. 1 of Competition 21 and Problem A.M. 1 of Competition 22.",
  "vars": [
    "n",
    "x",
    "y",
    "f",
    "v_n"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexn",
        "x": "inputx",
        "y": "inputy",
        "f": "funcfx",
        "v_n": "vindexn"
      },
      "question": "15. Which is greater\n\\[\n(\\sqrt{indexn})^{\\sqrt{indexn+1}} \\text { or }(\\sqrt{indexn+1})^{\\sqrt{indexn}}\n\\]\nwhere \\( indexn>8 \\) ?",
      "solution": "Solution. \\( (\\sqrt{indexn})^{\\sqrt{indexn+1}} \\) is greater than \\( (\\sqrt{indexn+1})^{\\sqrt{indexn}} \\) for \\( indexn>8 \\). Consider the function \\( funcfx(inputx)=(\\log inputx) / inputx \\) for \\( inputx>0 \\). Its derivative is \\( (1-\\log inputx) / inputx^{2} \\) which is negative for \\( inputx>e \\).\n\nHence, if \\( e \\leq inputx<inputy \\) we have \\( funcfx(inputx)>funcfx(inputy) \\), and\n\\[\ninputx inputy\\left(\\frac{\\log inputx}{inputx}\\right)>inputx inputy\\left(\\frac{\\log inputy}{inputy}\\right) .\n\\]\n\nTaking exponentials we get\n\\[\n\\boldsymbol{e}^{inputy \\log inputx}>\\boldsymbol{e}^{inputx \\log inputy},\n\\]\nthat is,\n\\[\ninputx^{inputy}>inputy^{inputx}\n\\]\nprovided \\( e \\leq inputx<inputy \\).\nIf \\( indexn \\geq 8 \\), then \\( e<\\sqrt{indexn}<\\sqrt{indexn+1} \\), so\n\\[\n(\\sqrt{indexn})^{\\sqrt{indexn+1}}>(\\sqrt{indexn+1})^{vindexn} .\n\\]\n\nRemark. A number of interesting problems are based on the inequality (1). See for example Journal of Recreational Mathematics, vol. 2, no. 4 (October 1969), pages 255-256, where the inequality \\( e^{\\pi}>\\pi^{e} \\) and some generalizations are discussed. The inequality is related to Problem P.M. 1 of Competition 21 and Problem A.M. 1 of Competition 22."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "zephyrion",
        "x": "pandorium",
        "y": "quasarine",
        "f": "nebulance",
        "v_n": "astronauta"
      },
      "question": "15. Which is greater\n\\[\n(\\sqrt{zephyrion})^{\\sqrt{zephyrion+1}} \\text { or }(\\sqrt{zephyrion+1})^{\\sqrt{zephyrion}}\n\\]\nwhere \\( zephyrion>8 \\) ?",
      "solution": "Solution. \\( (\\sqrt{zephyrion})^{\\sqrt{zephyrion+1}} \\) is greater than \\( (\\sqrt{zephyrion+1})^{\\sqrt{zephyrion}} \\) for \\( zephyrion>8 \\). Consider the function \\( nebulance(pandorium)=(\\log pandorium) / pandorium \\) for \\( pandorium>0 \\). Its derivative is \\( (1-\\log pandorium) / pandorium^{2} \\) which is negative for \\( pandorium>e \\).\n\nHence, if \\( e \\leq pandorium<quasarine \\) we have \\( nebulance(pandorium)>nebulance(quasarine) \\), and\n\\[\npandorium\\, quasarine\\left(\\frac{\\log pandorium}{pandorium}\\right)>pandorium\\, quasarine\\left(\\frac{\\log quasarine}{quasarine}\\right) .\n\\]\n\nTaking exponentials we get\n\\[\n\\boldsymbol{e}^{quasarine \\log pandorium}>\\boldsymbol{e}^{pandorium \\log quasarine},\n\\]\nthat is,\n\\[\npandorium^{quasarine}>quasarine^{pandorium}\n\\]\nprovided \\( e \\leq pandorium<quasarine \\).\nIf \\( zephyrion \\geq 8 \\), then \\( e<\\sqrt{zephyrion}<\\sqrt{zephyrion+1} \\), so\n\\[\n(\\sqrt{zephyrion})^{\\sqrt{zephyrion+1}}>(\\sqrt{zephyrion+1})^{astronauta} .\n\\]\n\nRemark. A number of interesting problems are based on the inequality (1). See for example Journal of Recreational Mathematics, vol. 2, no. 4 (October 1969), pages 255-256, where the inequality \\( e^{\\pi}>\\pi^{e} \\) and some generalizations are discussed. The inequality is related to Problem P.M. 1 of Competition 21 and Problem A.M. 1 of Competition 22."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "endlesscount",
        "x": "steadyvalue",
        "y": "lesserpoint",
        "f": "staticentity",
        "v_n": "firmscalar"
      },
      "question": "15. Which is greater\n\\[\n(\\sqrt{endlesscount})^{\\sqrt{endlesscount+1}} \\text { or }(\\sqrt{endlesscount+1})^{\\sqrt{endlesscount}}\n\\]\nwhere \\( endlesscount>8 \\) ?",
      "solution": "Solution. \\( (\\sqrt{endlesscount})^{\\sqrt{endlesscount+1}} \\) is greater than \\( (\\sqrt{endlesscount+1})^{\\sqrt{endlesscount}} \\) for \\( endlesscount>8 \\). Consider the function \\( staticentity(steadyvalue)=(\\log steadyvalue) / steadyvalue \\) for \\( steadyvalue>0 \\). Its derivative is \\( (1-\\log steadyvalue) / steadyvalue^{2} \\) which is negative for \\( steadyvalue>e \\).\n\nHence, if \\( e \\leq steadyvalue<lesserpoint \\) we have \\( staticentity(steadyvalue)>staticentity(lesserpoint) \\), and\n\\[\nsteadyvalue lesserpoint\\left(\\frac{\\log steadyvalue}{steadyvalue}\\right)>steadyvalue lesserpoint\\left(\\frac{\\log lesserpoint}{lesserpoint}\\right) .\n\\]\n\nTaking exponentials we get\n\\[\n\\boldsymbol{e}^{lesserpoint \\log steadyvalue}>\\boldsymbol{e}^{steadyvalue \\log lesserpoint},\n\\]\nthat is,\n\\[\nsteadyvalue^{lesserpoint}>lesserpoint^{steadyvalue}\n\\]\nprovided \\( e \\leq steadyvalue<lesserpoint \\).\nIf \\( endlesscount \\geq 8 \\), then \\( e<\\sqrt{endlesscount}<\\sqrt{endlesscount+1} \\), so\n\\[\n(\\sqrt{endlesscount})^{\\sqrt{endlesscount+1}}>(\\sqrt{endlesscount+1})^{firmscalar} .\n\\]\n\nRemark. A number of interesting problems are based on the inequality (1). See for example Journal of Recreational Mathematics, vol. 2, no. 4 (October 1969), pages 255-256, where the inequality \\( e^{\\pi}>\\pi^{e} \\) and some generalizations are discussed. The inequality is related to Problem P.M. 1 of Competition 21 and Problem A.M. 1 of Competition 22."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "x": "hjgrksla",
        "y": "nmbcvlqe",
        "f": "zlxksmpt",
        "v_n": "bgtrplsw"
      },
      "question": "15. Which is greater\n\\[\n(\\sqrt{qzxwvtnp})^{\\sqrt{qzxwvtnp+1}} \\text { or }(\\sqrt{qzxwvtnp+1})^{\\sqrt{qzxwvtnp}}\n\\]\nwhere \\( qzxwvtnp>8 \\) ?",
      "solution": "Solution. \\( (\\sqrt{qzxwvtnp})^{\\sqrt{qzxwvtnp+1}} \\) is greater than \\( (\\sqrt{qzxwvtnp+1})^{\\sqrt{qzxwvtnp}} \\) for \\( qzxwvtnp>8 \\). Consider the function \\( zlxksmpt(hjgrksla)=(\\log hjgrksla) / hjgrksla \\) for \\( hjgrksla>0 \\). Its derivative is \\( (1-\\log hjgrksla) / hjgrksla^{2} \\) which is negative for \\( hjgrksla>e \\).\n\nHence, if \\( e \\leq hjgrksla<nmbcvlqe \\) we have \\( zlxksmpt(hjgrksla)>zlxksmpt(nmbcvlqe) \\), and\n\\[\nhjgrksla nmbcvlqe\\left(\\frac{\\log hjgrksla}{hjgrksla}\\right)>hjgrksla nmbcvlqe\\left(\\frac{\\log nmbcvlqe}{nmbcvlqe}\\right) .\n\\]\n\nTaking exponentials we get\n\\[\n\\boldsymbol{e}^{nmbcvlqe \\log hjgrksla}>\\boldsymbol{e}^{hjgrksla \\log nmbcvlqe},\n\\]\nthat is,\n\\[\nhjgrksla^{nmbcvlqe}>nmbcvlqe^{hjgrksla}\n\\]\nprovided \\( e \\leq hjgrksla<nmbcvlqe \\).\nIf \\( qzxwvtnp \\geq 8 \\), then \\( e<\\sqrt{qzxwvtnp}<\\sqrt{qzxwvtnp+1} \\), so\n\\[\n(\\sqrt{qzxwvtnp})^{\\sqrt{qzxwvtnp+1}}>(\\sqrt{qzxwvtnp+1})^{bgtrplsw} .\n\\]\n\nRemark. A number of interesting problems are based on the inequality (1). See for example Journal of Recreational Mathematics, vol. 2, no. 4 (October 1969), pages 255-256, where the inequality \\( e^{\\pi}>\\pi^{e} \\) and some generalizations are discussed. The inequality is related to Problem P.M. 1 of Competition 21 and Problem A.M. 1 of Competition 22."
    },
    "kernel_variant": {
      "question": "Fix an integer  \n  m ≥ 2.  \nFor every integer  \n  n ≥ N(m) := ⌈e^{\\,m+1}⌉  \nconsider the two positive numbers  \n\n  A_{n,m}=∏_{j=0}^{m} (n+j)^{\\,(n+j+1)^{\\,1/(m+1)}},  \n\n  B_{n,m}=∏_{j=0}^{m} (n+j+1)^{\\,(n+j)^{\\,1/(m+1)}}.  \n\nDetermine which of the two numbers A_{n,m} and B_{n,m} is larger.",
      "solution": "Throughout the proof put  \n\n c := 1/(m+1)  (so 0 < c ≤ 1/3 because m ≥ 2).                        (0)\n\nStep 1. Reduce the comparison to logarithms.  \nDefine  \n\n Δ_{n,m} := log A_{n,m} − log B_{n,m}.  \n\nA direct expansion gives  \n\n Δ_{n,m}=Σ_{j=0}^{m}\\!\\Bigl[(n+j+1)^{c}\\log(n+j)−(n+j)^{c}\\log(n+j+1)\\Bigr].  (1)\n\nThus A_{n,m}>B_{n,m} ⇔ Δ_{n,m}>0.\n\nStep 2. Introduce a decreasing auxiliary function.  \nLet  \n\n f(x):= log x / x^{c},  x>0.                                            (2)\n\nIts derivative is  \n\n f '(x)=x^{−1−c}(1−c log x).                                           (3)\n\nBecause c = 1/(m+1), we have 1−c log x<0 precisely when log x>1/c = m+1.  \nHence f is strictly decreasing on (e^{\\,m+1},∞).\n\nStep 3. A point-wise inequality.  \nFor every x>e^{\\,m+1}, monotonicity gives  \n\n f(x) > f(x+1).                                                        (4)\n\nMultiplying both sides of (4) by the positive factor x^{c}(x+1)^{c} yields  \n\n (x+1)^{c}\\log x > x^{c}\\log(x+1).                                    (5)\n\nDefine  \n\n g(x):=(x+1)^{c}\\log x−x^{c}\\log(x+1).                                (6)\n\nThen g(x)>0 for every x>e^{\\,m+1}.\n\nStep 4. Apply the point-wise result to every summand.  \nBecause n ≥ ⌈e^{\\,m+1}⌉ and e^{\\,m+1} is irrational (and thus not an integer), we actually have the strict inequality n>e^{\\,m+1}.  \nConsequently n+j>e^{\\,m+1} for every 0≤j≤m, so by (6)\n\n g(n+j)>0 for all j.                                                   (7)\n\nBut g(n+j) is exactly the j-th term in (1). Hence every summand of Δ_{n,m} is positive, and\n\n Δ_{n,m}=Σ_{j=0}^{m}g(n+j)>0.                                         (8)\n\nStep 5. Conclusion.  \nInequality (8) implies log A_{n,m}>log B_{n,m}; exponentiating gives\n\n A_{n,m}>B_{n,m} for every integer n ≥ N(m)=⌈e^{\\,m+1}⌉.              (9)\n\nTherefore the required comparison is settled:\n\n  boxed{\\,A_{n,m}\\;>\\;B_{n,m}\\text{ for all integers }n\\ge\\lceil e^{\\,m+1}\\rceil,\\,m\\ge2.\\,}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.379288",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension / more variables  \n   • The original compares two simple numbers, the new problem involves an arbitrary integer parameter $m$ and a product of $m+1$ intertwined factors.  \n   • The threshold $N(m)$ must be located explicitly in terms of $m$.\n\n2. Additional constraints  \n   • The comparison must hold uniformly for all $n\\ge N(m)$ and all fixed $m\\ge2$, not merely for large $n$ in one special case.\n\n3. More sophisticated structures  \n   • The proof requires analysing a function $f(x)=\\log x/x^{1/(m+1)}$, establishing monotonicity via calculus, and then converting that analytic property into a combinatorial–symmetry argument over $m+1$ indices.\n\n4. Deeper theoretical requirements  \n   • One must manage decreasing functions on unbounded domains, determine precise regions of monotonicity, and handle delicate pairwise cancellations in a non-trivial sum.\n\n5. Multiple interacting concepts  \n   • Calculus (derivatives and monotonicity), exponential–logarithmic inequalities, index symmetry, and careful bounding all interact.  \n   • The solution scales with $m$, demanding an argument that works in every dimension rather than a one-off trick.\n\nAll these layers make the enhanced variant markedly harder than both the original problem and the existing kernel variant, which dealt with a single fixed fractional power and required only a one-line monotonicity observation."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer  \n  m ≥ 2.  \nFor every integer  \n  n ≥ N(m) := ⌈e^{\\,m+1}⌉  \nconsider the two positive numbers  \n\n  A_{n,m}=∏_{j=0}^{m} (n+j)^{\\,(n+j+1)^{\\,1/(m+1)}},  \n\n  B_{n,m}=∏_{j=0}^{m} (n+j+1)^{\\,(n+j)^{\\,1/(m+1)}}.  \n\nDetermine which of the two numbers A_{n,m} and B_{n,m} is larger.",
      "solution": "Throughout the proof put  \n\n c := 1/(m+1)  (so 0 < c ≤ 1/3 because m ≥ 2).                        (0)\n\nStep 1. Reduce the comparison to logarithms.  \nDefine  \n\n Δ_{n,m} := log A_{n,m} − log B_{n,m}.  \n\nA direct expansion gives  \n\n Δ_{n,m}=Σ_{j=0}^{m}\\!\\Bigl[(n+j+1)^{c}\\log(n+j)−(n+j)^{c}\\log(n+j+1)\\Bigr].  (1)\n\nThus A_{n,m}>B_{n,m} ⇔ Δ_{n,m}>0.\n\nStep 2. Introduce a decreasing auxiliary function.  \nLet  \n\n f(x):= log x / x^{c},  x>0.                                            (2)\n\nIts derivative is  \n\n f '(x)=x^{−1−c}(1−c log x).                                           (3)\n\nBecause c = 1/(m+1), we have 1−c log x<0 precisely when log x>1/c = m+1.  \nHence f is strictly decreasing on (e^{\\,m+1},∞).\n\nStep 3. A point-wise inequality.  \nFor every x>e^{\\,m+1}, monotonicity gives  \n\n f(x) > f(x+1).                                                        (4)\n\nMultiplying both sides of (4) by the positive factor x^{c}(x+1)^{c} yields  \n\n (x+1)^{c}\\log x > x^{c}\\log(x+1).                                    (5)\n\nDefine  \n\n g(x):=(x+1)^{c}\\log x−x^{c}\\log(x+1).                                (6)\n\nThen g(x)>0 for every x>e^{\\,m+1}.\n\nStep 4. Apply the point-wise result to every summand.  \nBecause n ≥ ⌈e^{\\,m+1}⌉ and e^{\\,m+1} is irrational (and thus not an integer), we actually have the strict inequality n>e^{\\,m+1}.  \nConsequently n+j>e^{\\,m+1} for every 0≤j≤m, so by (6)\n\n g(n+j)>0 for all j.                                                   (7)\n\nBut g(n+j) is exactly the j-th term in (1). Hence every summand of Δ_{n,m} is positive, and\n\n Δ_{n,m}=Σ_{j=0}^{m}g(n+j)>0.                                         (8)\n\nStep 5. Conclusion.  \nInequality (8) implies log A_{n,m}>log B_{n,m}; exponentiating gives\n\n A_{n,m}>B_{n,m} for every integer n ≥ N(m)=⌈e^{\\,m+1}⌉.              (9)\n\nTherefore the required comparison is settled:\n\n  boxed{\\,A_{n,m}\\;>\\;B_{n,m}\\text{ for all integers }n\\ge\\lceil e^{\\,m+1}\\rceil,\\,m\\ge2.\\,}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.326996",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension / more variables  \n   • The original compares two simple numbers, the new problem involves an arbitrary integer parameter $m$ and a product of $m+1$ intertwined factors.  \n   • The threshold $N(m)$ must be located explicitly in terms of $m$.\n\n2. Additional constraints  \n   • The comparison must hold uniformly for all $n\\ge N(m)$ and all fixed $m\\ge2$, not merely for large $n$ in one special case.\n\n3. More sophisticated structures  \n   • The proof requires analysing a function $f(x)=\\log x/x^{1/(m+1)}$, establishing monotonicity via calculus, and then converting that analytic property into a combinatorial–symmetry argument over $m+1$ indices.\n\n4. Deeper theoretical requirements  \n   • One must manage decreasing functions on unbounded domains, determine precise regions of monotonicity, and handle delicate pairwise cancellations in a non-trivial sum.\n\n5. Multiple interacting concepts  \n   • Calculus (derivatives and monotonicity), exponential–logarithmic inequalities, index symmetry, and careful bounding all interact.  \n   • The solution scales with $m$, demanding an argument that works in every dimension rather than a one-off trick.\n\nAll these layers make the enhanced variant markedly harder than both the original problem and the existing kernel variant, which dealt with a single fixed fractional power and required only a one-line monotonicity observation."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}