{
  "index": "1951-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "\\begin{array}{l}\n\\text { 3. Show that if } x \\text { is positive, then }\\\\\n\\log _{e}(1+1 / x)>1^{\\prime}(1+x)\n\\end{array}",
  "solution": "First Solution.\n\\[\n\\log \\left(1+\\frac{1}{x}\\right)=\\int_{x}^{1+x} \\frac{d t}{t}>\\int_{x}^{1+x} \\frac{d t}{1+x}=\\frac{1}{1+x} .\n\\]\n\nSecond Solution. In the well-known inequality\n\\[\n\\log (1+y)<y \\quad \\text { for }-1<y<0\n\\]\nput \\( y=-1 /(1+x) \\). We obtain\n\\[\n\\log \\left(\\frac{x}{1+x}\\right)<-\\frac{1}{1+x}\n\\]\nand, on changing signs,\n\\[\n\\log \\left(1+\\frac{1}{x}\\right)>\\frac{1}{1+x}\n\\]",
  "vars": [
    "x",
    "t",
    "y"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "unknownx",
        "t": "dummyvar",
        "y": "auxiliary"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Show that if } unknownx \\text { is positive, then }\\\\\n\\log _{e}(1+1 / unknownx)>1^{\\prime}(1+unknownx)\n\\end{array}",
      "solution": "First Solution.\n\\[\n\\log \\left(1+\\frac{1}{unknownx}\\right)=\\int_{unknownx}^{1+unknownx} \\frac{d dummyvar}{dummyvar}>\\int_{unknownx}^{1+unknownx} \\frac{d dummyvar}{1+unknownx}=\\frac{1}{1+unknownx} .\n\\]\n\nSecond Solution. In the well-known inequality\n\\[\n\\log (1+auxiliary)<auxiliary \\quad \\text { for }-1<auxiliary<0\n\\]\nput \\( auxiliary=-1 /(1+unknownx) \\). We obtain\n\\[\n\\log \\left(\\frac{unknownx}{1+unknownx}\\right)<-\\frac{1}{1+unknownx}\n\\]\nand, on changing signs,\n\\[\n\\log \\left(1+\\frac{1}{unknownx}\\right)>\\frac{1}{1+unknownx}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "blueberry",
        "t": "sailboat",
        "y": "cactusarm"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Show that if } blueberry \\text { is positive, then }\\\\\n\\log _{e}(1+1 / blueberry)>1^{\\prime}(1+blueberry)\n\\end{array}",
      "solution": "First Solution.\n\\[\n\\log \\left(1+\\frac{1}{blueberry}\\right)=\\int_{blueberry}^{1+blueberry} \\frac{d sailboat}{sailboat}>\\int_{blueberry}^{1+blueberry} \\frac{d sailboat}{1+blueberry}=\\frac{1}{1+blueberry} .\n\\]\n\nSecond Solution. In the well-known inequality\n\\[\n\\log (1+cactusarm)<cactusarm \\quad \\text { for }-1<cactusarm<0\n\\]\nput \\( cactusarm=-1 /(1+blueberry) \\). We obtain\n\\[\n\\log \\left(\\frac{blueberry}{1+blueberry}\\right)<-\\frac{1}{1+blueberry}\n\\]\nand, on changing signs,\n\\[\n\\log \\left(1+\\frac{1}{blueberry}\\right)>\\frac{1}{1+blueberry}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "negativevalue",
        "t": "timelessness",
        "y": "steadiness"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Show that if } negativevalue \\text { is positive, then }\\\\\n\\log _{e}(1+1 / negativevalue)>1^{\\prime}(1+negativevalue)\n\\end{array}",
      "solution": "First Solution.\n\\[\n\\log \\left(1+\\frac{1}{negativevalue}\\right)=\\int_{negativevalue}^{1+negativevalue} \\frac{d timelessness}{timelessness}>\\int_{negativevalue}^{1+negativevalue} \\frac{d timelessness}{1+negativevalue}=\\frac{1}{1+negativevalue} .\n\\]\n\nSecond Solution. In the well-known inequality\n\\[\n\\log (1+steadiness)<steadiness \\quad \\text { for }-1<steadiness<0\n\\]\nput \\( steadiness=-1 /(1+negativevalue) \\). We obtain\n\\[\n\\log \\left(\\frac{negativevalue}{1+negativevalue}\\right)<-\\frac{1}{1+negativevalue}\n\\]\nand, on changing signs,\n\\[\n\\log \\left(1+\\frac{1}{negativevalue}\\right)>\\frac{1}{1+negativevalue}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "y": "mkldpqrw"
      },
      "question": "\\begin{array}{l}\n\\text { 3. Show that if } qzxwvtnp \\text { is positive, then }\\\\\n\\log _{e}(1+1 / qzxwvtnp)>1^{\\prime}(1+qzxwvtnp)\n\\end{array}",
      "solution": "First Solution.\n\\[\n\\log \\left(1+\\frac{1}{qzxwvtnp}\\right)=\\int_{qzxwvtnp}^{1+qzxwvtnp} \\frac{d hjgrksla}{hjgrksla}>\\int_{qzxwvtnp}^{1+qzxwvtnp} \\frac{d hjgrksla}{1+qzxwvtnp}=\\frac{1}{1+qzxwvtnp} .\n\\]\n\nSecond Solution. In the well-known inequality\n\\[\n\\log (1+mkldpqrw)<mkldpqrw \\quad \\text { for }-1<mkldpqrw<0\n\\]\nput \\( mkldpqrw=-\\frac{1}{1+qzxwvtnp} \\). We obtain\n\\[\n\\log \\left(\\frac{qzxwvtnp}{1+qzxwvtnp}\\right)<-\\frac{1}{1+qzxwvtnp}\n\\]\nand, on changing signs,\n\\[\n\\log \\left(1+\\frac{1}{qzxwvtnp}\\right)>\\frac{1}{1+qzxwvtnp}\n\\]"
    },
    "kernel_variant": {
      "question": "Fix an integer $m\\ge 3$ and an integer $r$ with $2\\le r\\le m$.  \nFor every real number $x>0$ define  \n\\[\nR_{r}(x)=m^{(r)}\\int_{0}^{1}\\frac{t^{\\,r-1}}{(x+mt)^{\\,r}}\\;dt,\n\\qquad (\\dagger)\n\\]\nwhere $m^{(j)}=m(m-1)\\cdots(m-j+1)\\;(j\\ge 1)$ and $m^{(0)}=1$.\n\n(a) Prove the two–sided estimate  \n\\[\n0<R_{r}(x)<\\frac{m^{(r)}}{r\\,x^{\\,r}},\\qquad x>0.\n\\]\n\n(b) Put $G_{r}(x)=x^{\\,r}\\,R_{r}(x)$.  \nShow that $G_{r}$ is strictly increasing on $(0,\\infty)$ and that  \n\\[\n\\lim_{x\\to\\infty}G_{r}(x)=\\frac{m^{(r)}}{r}.\n\\]\n\n(c) Specialise to $r=2$.  Prove the refined inequality  \n\\[\n\\sum_{k=0}^{m-1}\\ln\\!\\left(1+\\frac{1}{x+k}\\right)>\n\\frac{m}{x+m}+\\frac{m^{2}}{2(x+m)^{2}},\\qquad x>0,\n\\tag{*}\n\\]\nand show that the constant $\\tfrac12$ is best possible, i.e.  \n\\[\n\\lim_{x\\to\\infty}(x+m)^{2}\\!\\left[\\sum_{k=0}^{m-1}\\ln\\!\\left(1+\\frac{1}{x+k}\\right)-\\frac{m}{x+m}\\right]=\\frac{m^{2}}{2}.\n\\]\n\n(The Euler–Maclaurin formula yields $(\\dagger)$; Bernoulli numbers appear only implicitly.)\n\n--------------------------------------------------------------------",
      "solution": "Throughout we fix $m\\ge 3$ and $2\\le r\\le m$.\n\n1.  The telescopic logarithmic sum  \n   \\[\n   S_m(x):=\\sum_{k=0}^{m-1}\\ln\\!\\left(1+\\frac{1}{x+k}\\right),\\qquad x>0,\n   \\]\n   satisfies  \n   \\[\n   S_m(x)=\\ln(x+m)-\\ln x\n          =m\\int_{0}^{1}\\frac{dt}{x+mt}.\n          \\tag{1}\n   \\]\n   Applying the Euler–Maclaurin formula termwise to $f(t)=1/(x+mt)$ produces\n   the identity $(\\dagger)$; we omit the classical derivation.\n\n2.  Proof of part (a)  \n   The integrand in $(\\dagger)$ is positive, hence $R_{r}(x)>0$.  \n   Since $x+mt\\ge x$ for $t\\in[0,1]$,\n   \\[\n      R_{r}(x)\n      <m^{(r)}\\int_{0}^{1}\\frac{t^{\\,r-1}}{x^{\\,r}}\\;dt\n      =\\frac{m^{(r)}}{r\\,x^{\\,r}},\\qquad x>0.\n   \\]\n\n3.  Monotonicity and limit in part (b)  \n   Differentiating $(\\dagger)$ under the integral sign gives\n   \\[\n   R_{r}'(x)= -r\\,m^{(r)}\\!\\int_{0}^{1}\n             \\frac{t^{\\,r-1}}{(x+mt)^{\\,r+1}}\\;dt<0.\n   \\tag{2}\n   \\]\n   For $G_{r}(x)=x^{\\,r}R_{r}(x)$ one obtains\n   \\[\n   G_{r}'(x)=r\\,x^{\\,r-1}R_{r}(x)+x^{\\,r}R_{r}'(x)\n           =r\\,m^{(r)}m\\,x^{\\,r-1}\\!\\int_{0}^{1}\n             \\frac{t^{\\,r}}{(x+mt)^{\\,r+1}}\\;dt>0,\n   \\]\n   so $G_{r}$ is strictly increasing.\n\n   For fixed $t\\in[0,1]$,\n   $(x+mt)^{-r}=x^{-r}(1+O(1/x))$ as $x\\to\\infty$.  \n   Substituting this in $(\\dagger)$ yields\n   \\[\n     R_{r}(x)=\\frac{m^{(r)}}{r\\,x^{\\,r}}+O\\!\\bigl(x^{-(r+1)}\\bigr),\n   \\]\n   hence\n   \\[\n     G_{r}(x)=\\frac{m^{(r)}}{r}+O\\!\\bigl(1/x\\bigr)\n              \\xrightarrow[x\\to\\infty]{}\\frac{m^{(r)}}{r}.\n   \\]\n\n4.  Preparations for part (c) ($r=2$)  \n   Formula $(\\dagger)$ becomes\n   \\[\n      R_{2}(x)=m(m-1)\\int_{0}^{1}\\frac{t}{(x+mt)^{2}}\\;dt.\n      \\tag{3}\n   \\]\n   Integrating by parts in (1) (choose $u=1/(x+mt)$, $dv=dt$) gives\n   \\[\n      S_m(x)=\\frac{m}{x+m}+m^{2}\\int_{0}^{1}\\frac{t}{(x+mt)^{2}}\\;dt.\n      \\tag{4}\n   \\]\n   Combining (3) and (4):\n   \\[\n      S_m(x)=\\frac{m}{x+m}+\\frac{m}{m-1}\\,R_{2}(x).\n      \\tag{5}\n   \\]\n\n5.  A concrete lower bound for $R_{2}$  \n   Because $0\\le t\\le 1$ implies $x+mt\\le x+m$, we have\n   \\[\n     \\frac{t}{(x+mt)^{2}}\\ge\\frac{t}{(x+m)^{2}}.\n   \\]\n   Inserting this into (3) entails\n   \\[\n     R_{2}(x)\\ge m(m-1)\\int_{0}^{1}\\frac{t}{(x+m)^{2}}\\;dt\n              =\\frac{m(m-1)}{2(x+m)^{2}}.\n      \\tag{6}\n   \\]\n\n6.  Proof of the refined inequality (*)  \n   From (5) and (6)\n   \\[\n      S_m(x)\\ge\\frac{m}{x+m}+\\frac{m}{m-1}\\cdot\n                \\frac{m(m-1)}{2(x+m)^{2}}\n              =\\frac{m}{x+m}+\\frac{m^{2}}{2(x+m)^{2}},\n   \\]\n   and strict inequality holds because the estimate in (6) is strict for\n   $t<1$. This proves (*).\n\n7.  Optimality of the constant $\\tfrac12$  \n   The asymptotic expansion of $R_{2}$ (special case of the result in\n   part (b)) is\n   \\[\n      R_{2}(x)=\\frac{m(m-1)}{2x^{2}}+O\\!\\bigl(x^{-3}\\bigr),\\qquad x\\to\\infty.\n   \\]\n   Hence, by (5),\n   \\[\n   S_m(x)-\\frac{m}{x+m}\n          =\\frac{m}{m-1}\\,R_{2}(x)\n          =\\frac{m^{2}}{2x^{2}}+O\\!\\bigl(x^{-3}\\bigr).\n   \\]\n   Multiplying by $(x+m)^{2}=x^{2}+2mx+m^{2}=x^{2}+O(x)$ we get\n   \\[\n     (x+m)^{2}\\!\\left[S_m(x)-\\frac{m}{x+m}\\right]\n     =\\frac{m^{2}}{2}+O\\!\\bigl(x^{-1}\\bigr)\n     \\xrightarrow[x\\to\\infty]{}\\frac{m^{2}}{2}.\n   \\]\n   Therefore the coefficient $\\tfrac12$ in (*) is the largest constant,\n   independent of $x$ and $m$, that can stand in front of the second\n   term.\n\nParts (a)–(c) are thereby completely established.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.441412",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order control – The problem no longer asks merely for a first-order lower bound; it demands the full Euler–Maclaurin expansion up to order r−1 together with a sharp estimate for the remainder of order r.\n\n2. Sophisticated tools – A solution requires knowledge of Bernoulli numbers, falling factorials, the periodic Bernoulli polynomials, and the Euler–Maclaurin summation formula; none of these appear in the original problem.\n\n3. Sign-controlled remainder – One must keep precise track of alternating signs in the expansion and connect them with the parity of r, a subtlety completely absent from the original inequality.\n\n4. Multi-step argument – The proof invokes an integral representation, applies Euler–Maclaurin inside another integral, carefully evaluates boundary terms, bounds a non-trivial remainder, and analyses monotonicity through differentiation of a compound function. Each step must be executed and justified, making the chain of reasoning substantially longer and deeper.\n\n5. General parameter r – The statement subsumes infinitely many inequalities (r = 3,4,5,…) in one theorem; handling an arbitrary order simultaneously further escalates technical complexity."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer $m\\ge 3$ and an integer $r$ with $2\\le r\\le m$.  \nFor every real number $x>0$ define  \n\\[\nR_{r}(x)=m^{(r)}\\int_{0}^{1}\\frac{t^{\\,r-1}}{(x+mt)^{\\,r}}\\;dt,\n\\qquad (\\dagger)\n\\]\nwhere $m^{(j)}=m(m-1)\\cdots(m-j+1)\\;(j\\ge 1)$ and $m^{(0)}=1$.\n\n(a) Prove the two–sided estimate  \n\\[\n0<R_{r}(x)<\\frac{m^{(r)}}{r\\,x^{\\,r}},\\qquad x>0.\n\\]\n\n(b) Put $G_{r}(x)=x^{\\,r}\\,R_{r}(x)$.  \nShow that $G_{r}$ is strictly increasing on $(0,\\infty)$ and that  \n\\[\n\\lim_{x\\to\\infty}G_{r}(x)=\\frac{m^{(r)}}{r}.\n\\]\n\n(c) Specialise to $r=2$.  Prove the refined inequality  \n\\[\n\\sum_{k=0}^{m-1}\\ln\\!\\left(1+\\frac{1}{x+k}\\right)>\n\\frac{m}{x+m}+\\frac{m^{2}}{2(x+m)^{2}},\\qquad x>0,\n\\tag{*}\n\\]\nand show that the constant $\\tfrac12$ is best possible, i.e.  \n\\[\n\\lim_{x\\to\\infty}(x+m)^{2}\\!\\left[\\sum_{k=0}^{m-1}\\ln\\!\\left(1+\\frac{1}{x+k}\\right)-\\frac{m}{x+m}\\right]=\\frac{m^{2}}{2}.\n\\]\n\n(The Euler–Maclaurin formula yields $(\\dagger)$; Bernoulli numbers appear only implicitly.)\n\n--------------------------------------------------------------------",
      "solution": "Throughout we fix $m\\ge 3$ and $2\\le r\\le m$.\n\n1.  The telescopic logarithmic sum  \n   \\[\n   S_m(x):=\\sum_{k=0}^{m-1}\\ln\\!\\left(1+\\frac{1}{x+k}\\right),\\qquad x>0,\n   \\]\n   satisfies  \n   \\[\n   S_m(x)=\\ln(x+m)-\\ln x\n          =m\\int_{0}^{1}\\frac{dt}{x+mt}.\n          \\tag{1}\n   \\]\n   Applying the Euler–Maclaurin formula termwise to $f(t)=1/(x+mt)$ produces\n   the identity $(\\dagger)$; we omit the classical derivation.\n\n2.  Proof of part (a)  \n   The integrand in $(\\dagger)$ is positive, hence $R_{r}(x)>0$.  \n   Since $x+mt\\ge x$ for $t\\in[0,1]$,\n   \\[\n      R_{r}(x)\n      <m^{(r)}\\int_{0}^{1}\\frac{t^{\\,r-1}}{x^{\\,r}}\\;dt\n      =\\frac{m^{(r)}}{r\\,x^{\\,r}},\\qquad x>0.\n   \\]\n\n3.  Monotonicity and limit in part (b)  \n   Differentiating $(\\dagger)$ under the integral sign gives\n   \\[\n   R_{r}'(x)= -r\\,m^{(r)}\\!\\int_{0}^{1}\n             \\frac{t^{\\,r-1}}{(x+mt)^{\\,r+1}}\\;dt<0.\n   \\tag{2}\n   \\]\n   For $G_{r}(x)=x^{\\,r}R_{r}(x)$ one obtains\n   \\[\n   G_{r}'(x)=r\\,x^{\\,r-1}R_{r}(x)+x^{\\,r}R_{r}'(x)\n           =r\\,m^{(r)}m\\,x^{\\,r-1}\\!\\int_{0}^{1}\n             \\frac{t^{\\,r}}{(x+mt)^{\\,r+1}}\\;dt>0,\n   \\]\n   so $G_{r}$ is strictly increasing.\n\n   For fixed $t\\in[0,1]$,\n   $(x+mt)^{-r}=x^{-r}(1+O(1/x))$ as $x\\to\\infty$.  \n   Substituting this in $(\\dagger)$ yields\n   \\[\n     R_{r}(x)=\\frac{m^{(r)}}{r\\,x^{\\,r}}+O\\!\\bigl(x^{-(r+1)}\\bigr),\n   \\]\n   hence\n   \\[\n     G_{r}(x)=\\frac{m^{(r)}}{r}+O\\!\\bigl(1/x\\bigr)\n              \\xrightarrow[x\\to\\infty]{}\\frac{m^{(r)}}{r}.\n   \\]\n\n4.  Preparations for part (c) ($r=2$)  \n   Formula $(\\dagger)$ becomes\n   \\[\n      R_{2}(x)=m(m-1)\\int_{0}^{1}\\frac{t}{(x+mt)^{2}}\\;dt.\n      \\tag{3}\n   \\]\n   Integrating by parts in (1) (choose $u=1/(x+mt)$, $dv=dt$) gives\n   \\[\n      S_m(x)=\\frac{m}{x+m}+m^{2}\\int_{0}^{1}\\frac{t}{(x+mt)^{2}}\\;dt.\n      \\tag{4}\n   \\]\n   Combining (3) and (4):\n   \\[\n      S_m(x)=\\frac{m}{x+m}+\\frac{m}{m-1}\\,R_{2}(x).\n      \\tag{5}\n   \\]\n\n5.  A concrete lower bound for $R_{2}$  \n   Because $0\\le t\\le 1$ implies $x+mt\\le x+m$, we have\n   \\[\n     \\frac{t}{(x+mt)^{2}}\\ge\\frac{t}{(x+m)^{2}}.\n   \\]\n   Inserting this into (3) entails\n   \\[\n     R_{2}(x)\\ge m(m-1)\\int_{0}^{1}\\frac{t}{(x+m)^{2}}\\;dt\n              =\\frac{m(m-1)}{2(x+m)^{2}}.\n      \\tag{6}\n   \\]\n\n6.  Proof of the refined inequality (*)  \n   From (5) and (6)\n   \\[\n      S_m(x)\\ge\\frac{m}{x+m}+\\frac{m}{m-1}\\cdot\n                \\frac{m(m-1)}{2(x+m)^{2}}\n              =\\frac{m}{x+m}+\\frac{m^{2}}{2(x+m)^{2}},\n   \\]\n   and strict inequality holds because the estimate in (6) is strict for\n   $t<1$. This proves (*).\n\n7.  Optimality of the constant $\\tfrac12$  \n   The asymptotic expansion of $R_{2}$ (special case of the result in\n   part (b)) is\n   \\[\n      R_{2}(x)=\\frac{m(m-1)}{2x^{2}}+O\\!\\bigl(x^{-3}\\bigr),\\qquad x\\to\\infty.\n   \\]\n   Hence, by (5),\n   \\[\n   S_m(x)-\\frac{m}{x+m}\n          =\\frac{m}{m-1}\\,R_{2}(x)\n          =\\frac{m^{2}}{2x^{2}}+O\\!\\bigl(x^{-3}\\bigr).\n   \\]\n   Multiplying by $(x+m)^{2}=x^{2}+2mx+m^{2}=x^{2}+O(x)$ we get\n   \\[\n     (x+m)^{2}\\!\\left[S_m(x)-\\frac{m}{x+m}\\right]\n     =\\frac{m^{2}}{2}+O\\!\\bigl(x^{-1}\\bigr)\n     \\xrightarrow[x\\to\\infty]{}\\frac{m^{2}}{2}.\n   \\]\n   Therefore the coefficient $\\tfrac12$ in (*) is the largest constant,\n   independent of $x$ and $m$, that can stand in front of the second\n   term.\n\nParts (a)–(c) are thereby completely established.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.381602",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-order control – The problem no longer asks merely for a first-order lower bound; it demands the full Euler–Maclaurin expansion up to order r−1 together with a sharp estimate for the remainder of order r.\n\n2. Sophisticated tools – A solution requires knowledge of Bernoulli numbers, falling factorials, the periodic Bernoulli polynomials, and the Euler–Maclaurin summation formula; none of these appear in the original problem.\n\n3. Sign-controlled remainder – One must keep precise track of alternating signs in the expansion and connect them with the parity of r, a subtlety completely absent from the original inequality.\n\n4. Multi-step argument – The proof invokes an integral representation, applies Euler–Maclaurin inside another integral, carefully evaluates boundary terms, bounds a non-trivial remainder, and analyses monotonicity through differentiation of a compound function. Each step must be executed and justified, making the chain of reasoning substantially longer and deeper.\n\n5. General parameter r – The statement subsumes infinitely many inequalities (r = 3,4,5,…) in one theorem; handling an arbitrary order simultaneously further escalates technical complexity."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}