{
  "index": "1999-B-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Let $f$ be a real function with a continuous third derivative such that $f(x),\nf'(x), f''(x), f'''(x)$ are positive for all $x$.  Suppose that\n$f'''(x)\\leq f(x)$ for all $x$.  Show that $f'(x)<2f(x)$ for all $x$.",
  "solution": "\\setcounter{equation}{0}\n(based on work by Daniel Stronger)\nWe make repeated use of the following fact: if $f$ is a differentiable function on all of\n$\\RR$, $\\lim_{x \\to -\\infty} f(x) \\geq 0$, and $f'(x) > 0$ for all $x \\in \\RR$, then\n$f(x) > 0$ for all $x \\in \\RR$. (Proof: if $f(y) < 0$ for some $x$, then $f(x)< f(y)$ for all\n$x<y$ since $f'>0$, but then $\\lim_{x \\to -\\infty} f(x) \\leq f(y) < 0$.)\n\nFrom the inequality $f'''(x) \\leq f(x)$ we obtain\n\\[\nf'' f'''(x) \\leq f''(x) f(x) < f''(x) f(x) + f'(x)^2\n\\]\nsince $f'(x)$ is positive. Applying the fact to the difference between the right and left sides,\nwe get\n\\begin{equation}\n\\frac{1}{2} (f''(x))^2 < f(x) f'(x).\n\\end{equation}\n\nOn the other hand, since $f(x)$ and $f'''(x)$ are both positive for all $x$,\nwe have\n\\[\n2f'(x) f''(x) < 2f'(x)f''(x) + 2f(x) f'''(x).\n\\]\nApplying the fact to the difference between the sides yields\n\\begin{equation}\nf'(x)^2 \\leq 2f(x) f''(x).\n\\end{equation}\nCombining (1) and (2), we obtain\n\\begin{align*}\n\\frac{1}{2} \\left( \\frac{f'(x)^2}{2f(x)} \\right)^2\n&< \\frac{1}{2} (f''(x))^2 \\\\\n&< f(x) f'(x),\n\\end{align*}\nor $(f'(x))^3 < 8 f(x)^3$. We conclude $f'(x) < 2f(x)$, as desired.\n\nNote: one can actually prove the result with a smaller constant in place of\n2, as follows. Adding $\\frac{1}{2} f'(x) f'''(x)$ to both sides\nof (1) and again invoking the original bound\n$f'''(x) \\leq f(x)$, we get\n\\begin{align*}\n\\frac{1}{2} [f'(x) f'''(x) + (f''(x))^2] &< f(x) f'(x) + \\frac{1}{2} f'(x) f'''(x) \\\\\n&\\leq \\frac{3}{2} f(x) f'(x).\n\\end{align*}\nApplying the fact again, we get\n\\[\n\\frac{1}{2} f'(x) f''(x) < \\frac{3}{4} f(x)^2.\n\\]\nMultiplying both sides by $f'(x)$ and applying the fact once more, we get\n\\[\n\\frac{1}{6} (f'(x))^3 < \\frac{1}{4} f(x)^3.\n\\]\nFrom this we deduce $f'(x) < (3/2)^{1/3} f(x) < 2f(x)$, as desired.\n\nI don't know what the best constant is, except that it is not less than 1\n(because $f(x) = e^x$ satisfies the given conditions).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "f"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "y": "another",
        "f": "function"
      },
      "question": "Let $function$ be a real function with a continuous third derivative such that $function(variable),\nfunction'(variable), function''(variable), function'''(variable)$ are positive for all $variable$.  Suppose that\n$function'''(variable)\\leq function(variable)$ for all $variable$.  Show that $function'(variable)<2function(variable)$ for all $variable$.",
      "solution": "\\setcounter{equation}{0}\n(based on work by Daniel Stronger)\nWe make repeated use of the following fact: if $function$ is a differentiable function on all of\n$\\RR$, $\\lim_{variable \\to -\\infty} function(variable) \\geq 0$, and $function'(variable) > 0$ for all $variable \\in \\RR$, then\n$function(variable) > 0$ for all $variable \\in \\RR$. (Proof: if $function(another) < 0$ for some $variable$, then $function(variable)< function(another)$ for all\n$variable<another$ since $function'>0$, but then $\\lim_{variable \\to -\\infty} function(variable) \\leq function(another) < 0$.)\n\nFrom the inequality $function'''(variable) \\leq function(variable)$ we obtain\n\\[\nfunction''(variable) function'''(variable) \\leq function''(variable) function(variable) < function''(variable) function(variable) + function'(variable)^2\n\\]\nsince $function'(variable)$ is positive. Applying the fact to the difference between the right and left sides,\nwe get\n\\begin{equation}\n\\frac{1}{2} (function''(variable))^2 < function(variable) function'(variable).\n\\end{equation}\n\nOn the other hand, since $function(variable)$ and $function'''(variable)$ are both positive for all $variable$,\nwe have\n\\[\n2function'(variable) function''(variable) < 2function'(variable)function''(variable) + 2function(variable) function'''(variable).\n\\]\nApplying the fact to the difference between the sides yields\n\\begin{equation}\nfunction'(variable)^2 \\leq 2function(variable) function''(variable).\n\\end{equation}\nCombining (1) and (2), we obtain\n\\begin{align*}\n\\frac{1}{2} \\left( \\frac{function'(variable)^2}{2function(variable)} \\right)^2\n&< \\frac{1}{2} (function''(variable))^2 \\\\\n&< function(variable) function'(variable),\n\\end{align*}\nor $(function'(variable))^3 < 8 function(variable)^3$. We conclude $function'(variable) < 2function(variable)$, as desired.\n\nNote: one can actually prove the result with a smaller constant in place of\n2, as follows. Adding $\\frac{1}{2} function'(variable) function'''(variable)$ to both sides\nof (1) and again invoking the original bound\n$function'''(variable) \\leq function(variable)$, we get\n\\begin{align*}\n\\frac{1}{2} [function'(variable) function'''(variable) + (function''(variable))^2] &< function(variable) function'(variable) + \\frac{1}{2} function'(variable) function'''(variable) \\\\\n&\\leq \\frac{3}{2} function(variable) function'(variable).\n\\end{align*}\nApplying the fact again, we get\n\\[\n\\frac{1}{2} function'(variable) function''(variable) < \\frac{3}{4} function(variable)^2.\n\\]\nMultiplying both sides by $function'(variable)$ and applying the fact once more, we get\n\\[\n\\frac{1}{6} (function'(variable))^3 < \\frac{1}{4} function(variable)^3.\n\\]\nFrom this we deduce $function'(variable) < (3/2)^{1/3} function(variable) < 2function(variable)$, as desired.\n\nI don't know what the best constant is, except that it is not less than 1\n(because $function(variable) = e^{variable}$ satisfies the given conditions)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "expedition",
        "f": "conundrum"
      },
      "question": "Let $conundrum$ be a real function with a continuous third derivative such that $conundrum(lighthouse), conundrum'(lighthouse), conundrum''(lighthouse), conundrum'''(lighthouse)$ are positive for all $lighthouse$.  Suppose that $conundrum'''(lighthouse)\\leq conundrum(lighthouse)$ for all $lighthouse$.  Show that $conundrum'(lighthouse)<2conundrum(lighthouse)$ for all $lighthouse$.",
      "solution": "\\setcounter{equation}{0}\n(based on work by Daniel Stronger)\nWe make repeated use of the following fact: if $conundrum$ is a differentiable function on all of $\\RR$, $\\lim_{lighthouse \\to -\\infty} conundrum(lighthouse) \\geq 0$, and $conundrum'(lighthouse) > 0$ for all $lighthouse \\in \\RR$, then $conundrum(lighthouse) > 0$ for all $lighthouse \\in \\RR$. (Proof: if $conundrum(expedition) < 0$ for some $lighthouse$, then $conundrum(lighthouse)< conundrum(expedition)$ for all $lighthouse<expedition$ since $conundrum'>0$, but then $\\lim_{lighthouse \\to -\\infty} conundrum(lighthouse) \\leq conundrum(expedition) < 0$.)\n\nFrom the inequality $conundrum'''(lighthouse) \\leq conundrum(lighthouse)$ we obtain\n\\[\nconundrum'' conundrum'''(lighthouse) \\leq conundrum''(lighthouse) conundrum(lighthouse) < conundrum''(lighthouse) conundrum(lighthouse) + conundrum'(lighthouse)^2\n\\]\nsince $conundrum'(lighthouse)$ is positive. Applying the fact to the difference between the right and left sides, we get\n\\begin{equation}\n\\frac{1}{2} (conundrum''(lighthouse))^2 < conundrum(lighthouse) conundrum'(lighthouse).\n\\end{equation}\n\nOn the other hand, since $conundrum(lighthouse)$ and $conundrum'''(lighthouse)$ are both positive for all $lighthouse$, we have\n\\[\n2conundrum'(lighthouse) conundrum''(lighthouse) < 2conundrum'(lighthouse)conundrum''(lighthouse) + 2conundrum(lighthouse) conundrum'''(lighthouse).\n\\]\nApplying the fact to the difference between the sides yields\n\\begin{equation}\nconundrum'(lighthouse)^2 \\leq 2conundrum(lighthouse) conundrum''(lighthouse).\n\\end{equation}\nCombining (1) and (2), we obtain\n\\begin{align*}\n\\frac{1}{2} \\left( \\frac{conundrum'(lighthouse)^2}{2conundrum(lighthouse)} \\right)^2 &< \\frac{1}{2} (conundrum''(lighthouse))^2 \\\\ &< conundrum(lighthouse) conundrum'(lighthouse),\n\\end{align*}\nor $(conundrum'(lighthouse))^3 < 8 conundrum(lighthouse)^3$. We conclude $conundrum'(lighthouse) < 2conundrum(lighthouse)$, as desired.\n\nNote: one can actually prove the result with a smaller constant in place of 2, as follows. Adding $\\frac{1}{2} conundrum'(lighthouse) conundrum'''(lighthouse)$ to both sides of (1) and again invoking the original bound $conundrum'''(lighthouse) \\leq conundrum(lighthouse)$, we get\n\\begin{align*}\n\\frac{1}{2} [conundrum'(lighthouse) conundrum'''(lighthouse) + (conundrum''(lighthouse))^2] &< conundrum(lighthouse) conundrum'(lighthouse) + \\frac{1}{2} conundrum'(lighthouse) conundrum'''(lighthouse) \\\\ &\\leq \\frac{3}{2} conundrum(lighthouse) conundrum'(lighthouse).\n\\end{align*}\nApplying the fact again, we get\n\\[\n\\frac{1}{2} conundrum'(lighthouse) conundrum''(lighthouse) < \\frac{3}{4} conundrum(lighthouse)^2.\n\\]\nMultiplying both sides by conundrum'(lighthouse) and applying the fact once more, we get\n\\[\n\\frac{1}{6} (conundrum'(lighthouse))^3 < \\frac{1}{4} conundrum(lighthouse)^3.\n\\]\nFrom this we deduce $conundrum'(lighthouse) < (3/2)^{1/3} conundrum(lighthouse) < 2conundrum(lighthouse)$, as desired.\n\nI don't know what the best constant is, except that it is not less than 1 (because $conundrum(lighthouse) = e^{lighthouse}$ satisfies the given conditions)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "nowherepoint",
        "y": "fixedpoint",
        "f": "constantvalue"
      },
      "question": "Let $constantvalue$ be a real function with a continuous third derivative such that $constantvalue(nowherepoint),\nconstantvalue'(nowherepoint), constantvalue''(nowherepoint), constantvalue'''(nowherepoint)$ are positive for all $nowherepoint$.  Suppose that\n$constantvalue'''(nowherepoint)\\leq constantvalue(nowherepoint)$ for all $nowherepoint$.  Show that $constantvalue'(nowherepoint)<2constantvalue(nowherepoint)$ for all $nowherepoint$.",
      "solution": "\\setcounter{equation}{0}\n(based on work by Daniel Stronger)\nWe make repeated use of the following fact: if $constantvalue$ is a differentiable function on all of\n$\\RR$, $\\lim_{nowherepoint \\to -\\infty} constantvalue(nowherepoint) \\geq 0$, and $constantvalue'(nowherepoint) > 0$ for all $nowherepoint \\in \\RR$, then\n$constantvalue(nowherepoint) > 0$ for all $nowherepoint \\in \\RR$. (Proof: if $constantvalue(fixedpoint) < 0$ for some $nowherepoint$, then $constantvalue(nowherepoint)< constantvalue(fixedpoint)$ for all\n$nowherepoint<fixedpoint$ since $constantvalue'>0$, but then $\\lim_{nowherepoint \\to -\\infty} constantvalue(nowherepoint) \\leq constantvalue(fixedpoint) < 0$.)\n\nFrom the inequality $constantvalue'''(nowherepoint) \\leq constantvalue(nowherepoint)$ we obtain\n\\[\nconstantvalue'' constantvalue'''(nowherepoint) \\leq constantvalue''(nowherepoint) constantvalue(nowherepoint) < constantvalue''(nowherepoint) constantvalue(nowherepoint) + constantvalue'(nowherepoint)^2\n\\]\nsince $constantvalue'(nowherepoint)$ is positive. Applying the fact to the difference between the right and left sides,\nwe get\n\\begin{equation}\n\\frac{1}{2} (constantvalue''(nowherepoint))^2 < constantvalue(nowherepoint) constantvalue'(nowherepoint).\n\\end{equation}\n\nOn the other hand, since $constantvalue(nowherepoint)$ and $constantvalue'''(nowherepoint)$ are both positive for all $nowherepoint$,\nwe have\n\\[\n2constantvalue'(nowherepoint) constantvalue''(nowherepoint) < 2constantvalue'(nowherepoint)constantvalue''(nowherepoint) + 2constantvalue(nowherepoint) constantvalue'''(nowherepoint).\n\\]\nApplying the fact to the difference between the sides yields\n\\begin{equation}\nconstantvalue'(nowherepoint)^2 \\leq 2constantvalue(nowherepoint) constantvalue''(nowherepoint).\n\\end{equation}\nCombining (1) and (2), we obtain\n\\begin{align*}\n\\frac{1}{2} \\left( \\frac{constantvalue'(nowherepoint)^2}{2constantvalue(nowherepoint)} \\right)^2\n&< \\frac{1}{2} (constantvalue''(nowherepoint))^2 \\\\\n&< constantvalue(nowherepoint) constantvalue'(nowherepoint),\n\\end{align*}\nor $(constantvalue'(nowherepoint))^3 < 8 constantvalue(nowherepoint)^3$. We conclude $constantvalue'(nowherepoint) < 2constantvalue(nowherepoint)$, as desired.\n\nNote: one can actually prove the result with a smaller constant in place of\n2, as follows. Adding $\\frac{1}{2} constantvalue'(nowherepoint) constantvalue'''(nowherepoint)$ to both sides\nof (1) and again invoking the original bound\n$constantvalue'''(nowherepoint) \\leq constantvalue(nowherepoint)$, we get\n\\begin{align*}\n\\frac{1}{2} [constantvalue'(nowherepoint) constantvalue'''(nowherepoint) + (constantvalue''(nowherepoint))^2] &< constantvalue(nowherepoint) constantvalue'(nowherepoint) + \\frac{1}{2} constantvalue'(nowherepoint) constantvalue'''(nowherepoint) \\\\\n&\\leq \\frac{3}{2} constantvalue(nowherepoint) constantvalue'(nowherepoint).\n\\end{align*}\nApplying the fact again, we get\n\\[\n\\frac{1}{2} constantvalue'(nowherepoint) constantvalue''(nowherepoint) < \\frac{3}{4} constantvalue(nowherepoint)^2.\n\\]\nMultiplying both sides by $constantvalue'(nowherepoint)$ and applying the fact once more, we get\n\\[\n\\frac{1}{6} (constantvalue'(nowherepoint))^3 < \\frac{1}{4} constantvalue(nowherepoint)^3.\n\\]\nFrom this we deduce $constantvalue'(nowherepoint) < (3/2)^{1/3} constantvalue(nowherepoint) < 2constantvalue(nowherepoint)$, as desired.\n\nI don't know what the best constant is, except that it is not less than 1\n(because $constantvalue(nowherepoint) = e^{nowherepoint}$ satisfies the given conditions)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "f": "plmbrtqk"
      },
      "question": "Let $plmbrtqk$ be a real function with a continuous third derivative such that $plmbrtqk(qzxwvtnp),\nplmbrtqk'(qzxwvtnp), plmbrtqk''(qzxwvtnp), plmbrtqk'''(qzxwvtnp)$ are positive for all $qzxwvtnp$.  Suppose that\n$plmbrtqk'''(qzxwvtnp)\\leq plmbrtqk(qzxwvtnp)$ for all $qzxwvtnp$.  Show that $plmbrtqk'(qzxwvtnp)<2plmbrtqk(qzxwvtnp)$ for all $qzxwvtnp$.",
      "solution": "\\setcounter{equation}{0}\n(based on work by Daniel Stronger)\nWe make repeated use of the following fact: if $plmbrtqk$ is a differentiable function on all of\n$\\RR$, $\\lim_{qzxwvtnp \\to -\\infty} plmbrtqk(qzxwvtnp) \\geq 0$, and $plmbrtqk'(qzxwvtnp) > 0$ for all $qzxwvtnp \\in \\RR$, then\n$plmbrtqk(qzxwvtnp) > 0$ for all $qzxwvtnp \\in \\RR$. (Proof: if $plmbrtqk(hjgrksla) < 0$ for some $qzxwvtnp$, then $plmbrtqk(qzxwvtnp)< plmbrtqk(hjgrksla)$ for all\n$qzxwvtnp<hjgrksla$ since $plmbrtqk'>0$, but then $\\lim_{qzxwvtnp \\to -\\infty} plmbrtqk(qzxwvtnp) \\leq plmbrtqk(hjgrksla) < 0$.)\n\nFrom the inequality $plmbrtqk'''(qzxwvtnp) \\leq plmbrtqk(qzxwvtnp)$ we obtain\n\\[\nplmbrtqk''\\,plmbrtqk'''(qzxwvtnp) \\leq plmbrtqk''(qzxwvtnp)\\,plmbrtqk(qzxwvtnp) < plmbrtqk''(qzxwvtnp)\\,plmbrtqk(qzxwvtnp) + plmbrtqk'(qzxwvtnp)^2\n\\]\nsince $plmbrtqk'(qzxwvtnp)$ is positive. Applying the fact to the difference between the right and left sides,\nwe get\n\\begin{equation}\n\\frac{1}{2} (plmbrtqk''(qzxwvtnp))^2 < plmbrtqk(qzxwvtnp)\\,plmbrtqk'(qzxwvtnp).\n\\end{equation}\n\nOn the other hand, since $plmbrtqk(qzxwvtnp)$ and $plmbrtqk'''(qzxwvtnp)$ are both positive for all $qzxwvtnp$,\nwe have\n\\[\n2\\,plmbrtqk'(qzxwvtnp)\\,plmbrtqk''(qzxwvtnp) < 2\\,plmbrtqk'(qzxwvtnp)plmbrtqk''(qzxwvtnp) + 2\\,plmbrtqk(qzxwvtnp)\\,plmbrtqk'''(qzxwvtnp).\n\\]\nApplying the fact to the difference between the sides yields\n\\begin{equation}\nplmbrtqk'(qzxwvtnp)^2 \\leq 2\\,plmbrtqk(qzxwvtnp)\\,plmbrtqk''(qzxwvtnp).\n\\end{equation}\nCombining (1) and (2), we obtain\n\\begin{align*}\n\\frac{1}{2} \\left( \\frac{plmbrtqk'(qzxwvtnp)^2}{2\\,plmbrtqk(qzxwvtnp)} \\right)^2\n&< \\frac{1}{2} (plmbrtqk''(qzxwvtnp))^2 \\\\\n&< plmbrtqk(qzxwvtnp)\\,plmbrtqk'(qzxwvtnp),\n\\end{align*}\nor $(plmbrtqk'(qzxwvtnp))^3 < 8\\,plmbrtqk(qzxwvtnp)^3$. We conclude $plmbrtqk'(qzxwvtnp) < 2\\,plmbrtqk(qzxwvtnp)$, as desired.\n\nNote: one can actually prove the result with a smaller constant in place of\n2, as follows. Adding $\\frac{1}{2}\\,plmbrtqk'(qzxwvtnp)\\,plmbrtqk'''(qzxwvtnp)$ to both sides\nof (1) and again invoking the original bound\n$plmbrtqk'''(qzxwvtnp) \\leq plmbrtqk(qzxwvtnp)$, we get\n\\begin{align*}\n\\frac{1}{2} [\\,plmbrtqk'(qzxwvtnp)\\,plmbrtqk'''(qzxwvtnp) + (plmbrtqk''(qzxwvtnp))^2] &< plmbrtqk(qzxwvtnp)\\,plmbrtqk'(qzxwvtnp) + \\frac{1}{2}\\,plmbrtqk'(qzxwvtnp)\\,plmbrtqk'''(qzxwvtnp) \\\\\n&\\leq \\frac{3}{2}\\,plmbrtqk(qzxwvtnp)\\,plmbrtqk'(qzxwvtnp).\n\\end{align*}\nApplying the fact again, we get\n\\[\n\\frac{1}{2}\\,plmbrtqk'(qzxwvtnp)\\,plmbrtqk''(qzxwvtnp) < \\frac{3}{4}\\,plmbrtqk(qzxwvtnp)^2.\n\\]\nMultiplying both sides by $plmbrtqk'(qzxwvtnp)$ and applying the fact once more, we get\n\\[\n\\frac{1}{6}\\,(plmbrtqk'(qzxwvtnp))^3 < \\frac{1}{4}\\,plmbrtqk(qzxwvtnp)^3.\n\\]\nFrom this we deduce $plmbrtqk'(qzxwvtnp) < (3/2)^{1/3}\\,plmbrtqk(qzxwvtnp) < 2\\,plmbrtqk(qzxwvtnp)$, as desired.\n\nI don't know what the best constant is, except that it is not less than 1\n(because $plmbrtqk(qzxwvtnp) = e^{qzxwvtnp}$ satisfies the given conditions)."
    },
    "kernel_variant": {
      "question": "Let $f:\\mathbb R\\to(0,\\infty)$ be three times continuously differentiable and suppose\n\\[\n    f'(x),\\,f''(x),\\,f'''(x)\\ge 0\\quad(\\text{for every }x\\in\\mathbb R)\n\\]\nand\n\\[\n    f'''(x)\\le 3\\,f(x)\\quad(\\text{for every }x\\in\\mathbb R).\n\\]\nProve that\n\\[\n      f'(x)<2\\,3^{1/3}\\,f(x)\\qquad\\text{for all }x\\in\\mathbb R.\n\\]",
      "solution": "Let f∈C³(ℝ) satisfy f(x)>0, f′(x)≥0, f″(x)≥0, f‴(x)≥0 and f‴(x)≤3f(x) for all x.  We first record the correct monotonic–sign lemma:\n\nLemma.  If g is differentiable on ℝ, g′(x)≥0 for all x, and limₓ→−∞ g(x)≥0, then g(x)≥0 for all x.\nProof.  Since g′≥0, g is nondecreasing.  If g(y)<0 for some y, then for all x<y we have g(x)≤g(y)<0, contradicting limₓ→−∞ g(x)≥0.\n\nNext, since f′, f″, f‴≥0 and f″ is nondecreasing (because f‴≥0), each of f′(x), f″(x) has a finite limit ≥0 as x→−∞.  If, say, L:=limₓ→−∞ f′(x)>0, then for x≪0 we would have f′(x)>L/2>0 and hence\n    f(x)=f(0)−∫ₓ⁰ f′(t)\n→−∞ as x→−∞,\ncontradicting f>0.  Thus f′(−∞)=f″(−∞)=0, and similarly f(−∞)=limₓ→−∞ f(x) exists and ≥0.\n\n1. Define g₁(x)=½(f″(x))²−3f(x)f′(x).  Then\n   g₁′(x)=f″(x)f‴(x)−3(f′(x))²−3f(x)f″(x)\n          ≤3f(x)f″(x)−3(f′(x))²−3f(x)f″(x)\n          =−3(f′(x))²≤0.\nHence g₁ is nonincreasing, and limₓ→−∞ g₁(x)=½·0²−3·f(−∞)·0=0.  By the lemma applied to −g₁ (which is nondecreasing with limit ≥0 at −∞), we get −g₁(x)≥0, i.e.\n   ½(f″(x))²≤3f(x)f′(x).\n\n2. Define g₂(x)=(f′(x))²−2f(x)f″(x).  Then\n   g₂′(x)=2f′f″−2f′f″−2f f‴=−2f(x)f‴(x)≤0.\nSo g₂ is nonincreasing, limₓ→−∞ g₂(x)=0−0=0, and again by the lemma applied to −g₂ we obtain\n   (f′(x))²≤2f(x)f″(x).\n\n3. From (2) we have f″(x)≥(f′(x))²/[2f(x)].  Substitute into (1):\n   ½·((f′)²/(2f))²≤3f f′\n   ⇒f′⁴/(8f²)≤3f f′\n   ⇒f′³≤24f³.\nSince f′≥0 we take cube‐roots to conclude\n   f′(x)≤(24)^{1/3}f(x)=2·3^{1/3}f(x).\nMoreover, strict inequality holds wherever f′>0, and at points where f′=0 the bound is trivial.  Hence for all x,\n   f′(x)<2·3^{1/3}f(x),\nas required. ■",
      "_meta": {
        "core_steps": [
          "Monotone-positivity lemma: if g' > 0 and lim_{x→−∞} g ≥ 0 then g(x) > 0",
          "Apply lemma to g₁(x)=½(f''(x))² − f(x)f'(x) to get ½(f'')² < f f'",
          "Apply lemma to g₂(x)=f'(x)² − 2 f(x)f''(x) to get (f')² ≤ 2 f f''",
          "Combine the two inequalities to obtain (f')³ < 8 f³",
          "Conclude desired bound: f'(x) < 2 f(x)"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Coefficient in the hypothesis f'''(x) ≤ k·f(x)",
            "original": "k = 1"
          },
          "slot2": {
            "description": "Direction used in the limit for the auxiliary lemma (−∞ vs +∞)",
            "original": "lim_{x→−∞}"
          },
          "slot3": {
            "description": "Strict positivity of f, f', f'', f'''; can be relaxed to non-negativity with minor tweaks",
            "original": "f, f', f'', f''' > 0"
          },
          "slot4": {
            "description": "Final constant C such that f'(x) < C·f(x); equals 2 when k = 1",
            "original": "C = 2"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}