{
  "index": "1973-B-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-4. (a) On [0,1], let \\( f \\) have a continuous derivative satisfying \\( 0<f^{\\prime}(x) \\leqq 1 \\). Also suppose that \\( f(0)=0 \\). Prove that\n\\[\n\\left[\\int_{0}^{1} f(x) d x\\right]^{2} \\geqq \\int_{0}^{1}[f(x)]^{3} d x\n\\]\n[Hint: Replace the inequality by one involving the inverse function to \\( f \\).]\n(b) Show an example in which equality occurs.",
  "solution": "B-4. We give two solutions; the first does not use the hint and the second does.\nTheorem. If \\( f \\) is continuous on \\( [0,1], f(0)=0 \\), and \\( 0 \\leqq f^{\\prime}(x) \\leqq 1 \\) on \\( (0,1) \\), then\n\\[\n\\left[\\int_{0}^{1} f(x) d x\\right]^{2}>\\int_{0}^{1}[f(x)]^{3} d x\n\\]\nunless, identically on \\( [0,1] \\), either \\( f(x)=x \\) or \\( f(x)=0 \\).\nProof. Define \\( G(t)=2 \\int_{0}^{t} f(x) d x-[f(t)]^{2} \\) for \\( t \\in[0,1] \\). Then \\( G(0)=0 \\) and \\( G^{\\prime}(t)=2 f(t)\\left[1-f^{\\prime}(t)\\right] \\geqq 0 \\), so that \\( G(t) \\geqq 0 \\) and consequently \\( f(t) G(t) \\geqq 0 \\).\n\nNow define \\( F(t)=\\left[\\int_{0}^{t} f(x) d x\\right]^{2}-\\int_{0}^{t}[f(x)]^{3} d x \\) for \\( t \\in[0,1] \\). Then \\( F(0)=0 \\) and \\( F^{\\prime}(t)=f(t) G(t) \\geqq 0 \\), so that \\( F(t) \\geqq 0 \\) and in particular \\( F(1) \\geqq 0 \\).\n\nEquality is possible only if \\( f(t) G(t)=F^{\\prime}(t)=0 \\) for all \\( t \\), which implies that, for some \\( K, f=0 \\) on \\( [0, K] \\) and \\( G^{\\prime}=0 \\), with \\( f>0 \\), on \\( (K, 1) \\). We then have \\( f^{\\prime}=1 \\) on \\( (K, 1) \\), which is admissible only if \\( K=0 \\) or \\( K=1 \\), since otherwise \\( f^{\\prime}(K) \\) is simultaneously defined and undefined.\n\nThe unique answer to (b) is \\( f(x)=x \\). The following is an outline of a proof of (a) using the hint. Let \\( f(1)=c \\). The hypothesis implies that \\( f \\) has an inverse \\( g \\) with \\( g^{\\prime}(y) \\geqq 1 \\) on \\( 0 \\leqq y \\leqq c \\). Let\n\\[\nA=\\left[\\int_{0}^{1} f(x) d x\\right]^{2} \\text { and } B=\\int_{0}^{1}[f(x)]^{3} d x\n\\]\n\nThen\n\\[\nA=\\left[\\int_{0}^{c} y g^{\\prime}(y) d y\\right]^{2}=\\int_{0}^{c} \\int_{0}^{c} y g^{\\prime}(y) z g^{\\prime}(z) d z d y=2 \\int_{0}^{c} \\int_{0}^{z} y g^{\\prime}(y) z g^{\\prime}(z) d y d z\n\\]\nusing the symmetry of the integrand about the line \\( y=z \\). Now \\( g^{\\prime}(y) \\geqq 1 \\) implies\n\\[\nA \\geqq \\int_{0}^{c} z g^{\\prime}(z)\\left[\\int_{0}^{z} 2 y d y\\right] d z=\\int_{0}^{c} z^{3} g^{\\prime}(z) d z=B\n\\]",
  "vars": [
    "x",
    "t",
    "y",
    "z"
  ],
  "params": [
    "f",
    "g",
    "G",
    "F",
    "K",
    "c",
    "A",
    "B"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "domainx",
        "t": "paramt",
        "y": "coordy",
        "z": "coordz",
        "f": "mapfunc",
        "g": "invfunc",
        "G": "auxgfun",
        "F": "auxffun",
        "K": "splitpt",
        "c": "endpoint",
        "A": "integrala",
        "B": "integralb"
      },
      "question": "B-4. (a) On [0,1], let \\( mapfunc \\) have a continuous derivative satisfying \\( 0<mapfunc^{\\prime}(domainx) \\leqq 1 \\). Also suppose that \\( mapfunc(0)=0 \\). Prove that\n\\[\n\\left[\\int_{0}^{1} mapfunc(domainx) d domainx\\right]^{2} \\geqq \\int_{0}^{1}[mapfunc(domainx)]^{3} d domainx\n\\]\n[Hint: Replace the inequality by one involving the inverse function to \\( mapfunc \\).]\n(b) Show an example in which equality occurs.",
      "solution": "B-4. We give two solutions; the first does not use the hint and the second does.\nTheorem. If \\( mapfunc \\) is continuous on \\( [0,1], mapfunc(0)=0 \\), and \\( 0 \\leqq mapfunc^{\\prime}(domainx) \\leqq 1 \\) on \\( (0,1) \\), then\n\\[\n\\left[\\int_{0}^{1} mapfunc(domainx) d domainx\\right]^{2}>\\int_{0}^{1}[mapfunc(domainx)]^{3} d domainx\n\\]\nunless, identically on \\( [0,1] \\), either \\( mapfunc(domainx)=domainx \\) or \\( mapfunc(domainx)=0 \\).\nProof. Define \\( auxgfun(paramt)=2 \\int_{0}^{paramt} mapfunc(domainx) d domainx-[mapfunc(paramt)]^{2} \\) for \\( paramt \\in[0,1] \\). Then \\( auxgfun(0)=0 \\) and \\( auxgfun^{\\prime}(paramt)=2 mapfunc(paramt)\\left[1-mapfunc^{\\prime}(paramt)\\right] \\geqq 0 \\), so that \\( auxgfun(paramt) \\geqq 0 \\) and consequently \\( mapfunc(paramt) auxgfun(paramt) \\geqq 0 \\).\n\nNow define \\( auxffun(paramt)=\\left[\\int_{0}^{paramt} mapfunc(domainx) d domainx\\right]^{2}-\\int_{0}^{paramt}[mapfunc(domainx)]^{3} d domainx \\) for \\( paramt \\in[0,1] \\). Then \\( auxffun(0)=0 \\) and \\( auxffun^{\\prime}(paramt)=mapfunc(paramt) auxgfun(paramt) \\geqq 0 \\), so that \\( auxffun(paramt) \\geqq 0 \\) and in particular \\( auxffun(1) \\geqq 0 \\).\n\nEquality is possible only if \\( mapfunc(paramt) auxgfun(paramt)=auxffun^{\\prime}(paramt)=0 \\) for all \\( paramt \\), which implies that, for some \\( splitpt, mapfunc=0 \\) on \\( [0, splitpt] \\) and \\( auxgfun^{\\prime}=0 \\), with \\( mapfunc>0 \\), on \\( (splitpt, 1) \\). We then have \\( mapfunc^{\\prime}=1 \\) on \\( (splitpt, 1) \\), which is admissible only if \\( splitpt=0 \\) or \\( splitpt=1 \\), since otherwise \\( mapfunc^{\\prime}(splitpt) \\) is simultaneously defined and undefined.\n\nThe unique answer to (b) is \\( mapfunc(domainx)=domainx \\). The following is an outline of a proof of (a) using the hint. Let \\( mapfunc(1)=endpoint \\). The hypothesis implies that \\( mapfunc \\) has an inverse \\( invfunc \\) with \\( invfunc^{\\prime}(coordy) \\geqq 1 \\) on \\( 0 \\leqq coordy \\leqq endpoint \\). Let\n\\[\nintegrala=\\left[\\int_{0}^{1} mapfunc(domainx) d domainx\\right]^{2} \\text { and } integralb=\\int_{0}^{1}[mapfunc(domainx)]^{3} d domainx\n\\]\n\nThen\n\\[\nintegrala=\\left[\\int_{0}^{endpoint} coordy invfunc^{\\prime}(coordy) d coordy\\right]^{2}=\\int_{0}^{endpoint} \\int_{0}^{endpoint} coordy invfunc^{\\prime}(coordy) coordz invfunc^{\\prime}(coordz) d coordz d coordy=2 \\int_{0}^{endpoint} \\int_{0}^{coordz} coordy invfunc^{\\prime}(coordy) coordz invfunc^{\\prime}(coordz) d coordy d coordz\n\\]\nusing the symmetry of the integrand about the line \\( coordy=coordz \\). Now \\( invfunc^{\\prime}(coordy) \\geqq 1 \\) implies\n\\[\nintegrala \\geqq \\int_{0}^{endpoint} coordz invfunc^{\\prime}(coordz)\\left[\\int_{0}^{coordz} 2 coordy d coordy\\right] d coordz=\\int_{0}^{endpoint} coordz^{3} invfunc^{\\prime}(coordz) d coordz=integralb\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineconee",
        "t": "stormcloud",
        "y": "driftwood",
        "z": "riverstone",
        "f": "labyrinth",
        "g": "afterglow",
        "G": "sandstone",
        "F": "aquamarine",
        "K": "blacksmith",
        "c": "columbine",
        "A": "ambergris",
        "B": "bloodstone"
      },
      "question": "B-4. (a) On [0,1], let \\( labyrinth \\) have a continuous derivative satisfying \\( 0<labyrinth^{\\prime}(pineconee) \\leqq 1 \\). Also suppose that \\( labyrinth(0)=0 \\). Prove that\n\\[\n\\left[\\int_{0}^{1} labyrinth(pineconee) d pineconee\\right]^{2} \\geqq \\int_{0}^{1}[labyrinth(pineconee)]^{3} d pineconee\n\\]\n[Hint: Replace the inequality by one involving the inverse function to \\( labyrinth \\).]\n(b) Show an example in which equality occurs.",
      "solution": "B-4. We give two solutions; the first does not use the hint and the second does.\nTheorem. If \\( labyrinth \\) is continuous on \\( [0,1], labyrinth(0)=0 \\), and \\( 0 \\leqq labyrinth^{\\prime}(pineconee) \\leqq 1 \\) on \\( (0,1) \\), then\n\\[\n\\left[\\int_{0}^{1} labyrinth(pineconee) d pineconee\\right]^{2}>\\int_{0}^{1}[labyrinth(pineconee)]^{3} d pineconee\n\\]\nunless, identically on \\( [0,1] \\), either \\( labyrinth(pineconee)=pineconee \\) or \\( labyrinth(pineconee)=0 \\).\nProof. Define \\( sandstone(stormcloud)=2 \\int_{0}^{stormcloud} labyrinth(pineconee) d pineconee-[labyrinth(stormcloud)]^{2} \\) for \\( stormcloud \\in[0,1] \\). Then \\( sandstone(0)=0 \\) and \\( sandstone^{\\prime}(stormcloud)=2 labyrinth(stormcloud)\\left[1-labyrinth^{\\prime}(stormcloud)\\right] \\geqq 0 \\), so that \\( sandstone(stormcloud) \\geqq 0 \\) and consequently \\( labyrinth(stormcloud) sandstone(stormcloud) \\geqq 0 \\).\n\nNow define \\( aquamarine(stormcloud)=\\left[\\int_{0}^{stormcloud} labyrinth(pineconee) d pineconee\\right]^{2}-\\int_{0}^{stormcloud}[labyrinth(pineconee)]^{3} d pineconee \\) for \\( stormcloud \\in[0,1] \\). Then \\( aquamarine(0)=0 \\) and \\( aquamarine^{\\prime}(stormcloud)=labyrinth(stormcloud) sandstone(stormcloud) \\geqq 0 \\), so that \\( aquamarine(stormcloud) \\geqq 0 \\) and in particular \\( aquamarine(1) \\geqq 0 \\).\n\nEquality is possible only if \\( labyrinth(stormcloud) sandstone(stormcloud)=aquamarine^{\\prime}(stormcloud)=0 \\) for all \\( stormcloud \\), which implies that, for some \\( blacksmith, labyrinth=0 \\) on \\( [0, blacksmith] \\) and \\( sandstone^{\\prime}=0 \\), with \\( labyrinth>0 \\), on \\( (blacksmith, 1) \\). We then have \\( labyrinth^{\\prime}=1 \\) on \\( (blacksmith, 1) \\), which is admissible only if \\( blacksmith=0 \\) or \\( blacksmith=1 \\), since otherwise \\( labyrinth^{\\prime}(blacksmith) \\) is simultaneously defined and undefined.\n\nThe unique answer to (b) is \\( labyrinth(pineconee)=pineconee \\). The following is an outline of a proof of (a) using the hint. Let \\( labyrinth(1)=columbine \\). The hypothesis implies that \\( labyrinth \\) has an inverse \\( afterglow \\) with \\( afterglow^{\\prime}(driftwood) \\geqq 1 \\) on \\( 0 \\leqq driftwood \\leqq columbine \\). Let\n\\[\nambergris=\\left[\\int_{0}^{1} labyrinth(pineconee) d pineconee\\right]^{2} \\text { and } bloodstone=\\int_{0}^{1}[labyrinth(pineconee)]^{3} d pineconee\n\\]\n\nThen\n\\[\nambergris=\\left[\\int_{0}^{columbine} driftwood afterglow^{\\prime}(driftwood) d driftwood\\right]^{2}=\\int_{0}^{columbine} \\int_{0}^{columbine} driftwood afterglow^{\\prime}(driftwood) riverstone afterglow^{\\prime}(riverstone) d riverstone d driftwood=2 \\int_{0}^{columbine} \\int_{0}^{riverstone} driftwood afterglow^{\\prime}(driftwood) riverstone afterglow^{\\prime}(riverstone) d driftwood d riverstone\n\\]\nusing the symmetry of the integrand about the line \\( driftwood=riverstone \\). Now \\( afterglow^{\\prime}(driftwood) \\geqq 1 \\) implies\n\\[\nambergris \\geqq \\int_{0}^{columbine} riverstone afterglow^{\\prime}(riverstone)\\left[\\int_{0}^{riverstone} 2 driftwood d driftwood\\right] d riverstone=\\int_{0}^{columbine} riverstone^{3} afterglow^{\\prime}(riverstone) d riverstone=bloodstone\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "yonderconstant",
        "t": "spaceless",
        "y": "steadfast",
        "z": "frozenpoint",
        "f": "stationary",
        "g": "directmap",
        "G": "staticform",
        "F": "unchanging",
        "K": "fullspan",
        "c": "voidness",
        "A": "smallness",
        "B": "littleness"
      },
      "question": "B-4. (a) On [0,1], let \\( stationary \\) have a continuous derivative satisfying \\( 0<stationary^{\\prime}(yonderconstant) \\leqq 1 \\). Also suppose that \\( stationary(0)=0 \\). Prove that\n\\[\n\\left[\\int_{0}^{1} stationary(yonderconstant) d yonderconstant\\right]^{2} \\geqq \\int_{0}^{1}[stationary(yonderconstant)]^{3} d yonderconstant\n\\]\n[Hint: Replace the inequality by one involving the inverse function to \\( stationary \\).]\n(b) Show an example in which equality occurs.",
      "solution": "B-4. We give two solutions; the first does not use the hint and the second does.\nTheorem. If \\( stationary \\) is continuous on \\( [0,1], stationary(0)=0 \\), and \\( 0 \\leqq stationary^{\\prime}(yonderconstant) \\leqq 1 \\) on \\( (0,1) \\), then\n\\[\n\\left[\\int_{0}^{1} stationary(yonderconstant) d yonderconstant\\right]^{2}>\\int_{0}^{1}[stationary(yonderconstant)]^{3} d yonderconstant\n\\]\nunless, identically on \\( [0,1] \\), either \\( stationary(yonderconstant)=yonderconstant \\) or \\( stationary(yonderconstant)=0 \\).\nProof. Define \\( staticform(spaceless)=2 \\int_{0}^{spaceless} stationary(yonderconstant) d yonderconstant-[stationary(spaceless)]^{2} \\) for \\( spaceless \\in[0,1] \\). Then \\( staticform(0)=0 \\) and \\( staticform^{\\prime}(spaceless)=2 stationary(spaceless)\\left[1-stationary^{\\prime}(spaceless)\\right] \\geqq 0 \\), so that \\( staticform(spaceless) \\geqq 0 \\) and consequently \\( stationary(spaceless) staticform(spaceless) \\geqq 0 \\).\n\nNow define \\( unchanging(spaceless)=\\left[\\int_{0}^{spaceless} stationary(yonderconstant) d yonderconstant\\right]^{2}-\\int_{0}^{spaceless}[stationary(yonderconstant)]^{3} d yonderconstant \\) for \\( spaceless \\in[0,1] \\). Then \\( unchanging(0)=0 \\) and \\( unchanging^{\\prime}(spaceless)=stationary(spaceless) staticform(spaceless) \\geqq 0 \\), so that \\( unchanging(spaceless) \\geqq 0 \\) and in particular \\( unchanging(1) \\geqq 0 \\).\n\nEquality is possible only if \\( stationary(spaceless) staticform(spaceless)=unchanging^{\\prime}(spaceless)=0 \\) for all \\( spaceless \\), which implies that, for some \\( fullspan, stationary=0 \\) on \\( [0, fullspan] \\) and \\( staticform^{\\prime}=0 \\), with \\( stationary>0 \\), on \\( (fullspan, 1) \\). We then have \\( stationary^{\\prime}=1 \\) on \\( (fullspan, 1) \\), which is admissible only if \\( fullspan=0 \\) or \\( fullspan=1 \\), since otherwise \\( stationary^{\\prime}(fullspan) \\) is simultaneously defined and undefined.\n\nThe unique answer to (b) is \\( stationary(yonderconstant)=yonderconstant \\). The following is an outline of a proof of (a) using the hint. Let \\( stationary(1)=voidness \\). The hypothesis implies that \\( stationary \\) has an inverse \\( directmap \\) with \\( directmap^{\\prime}(steadfast) \\geqq 1 \\) on \\( 0 \\leqq steadfast \\leqq voidness \\). Let\n\\[\nsmallness=\\left[\\int_{0}^{1} stationary(yonderconstant) d yonderconstant\\right]^{2} \\text { and } littleness=\\int_{0}^{1}[stationary(yonderconstant)]^{3} d yonderconstant\n\\]\n\nThen\n\\[\nsmallness=\\left[\\int_{0}^{voidness} steadfast directmap^{\\prime}(steadfast) d steadfast\\right]^{2}=\\int_{0}^{voidness} \\int_{0}^{voidness} steadfast directmap^{\\prime}(steadfast) frozenpoint directmap^{\\prime}(frozenpoint) d frozenpoint d steadfast=2 \\int_{0}^{voidness} \\int_{0}^{frozenpoint} steadfast directmap^{\\prime}(steadfast) frozenpoint directmap^{\\prime}(frozenpoint) d steadfast d frozenpoint\n\\]\nusing the symmetry of the integrand about the line \\( steadfast=frozenpoint \\). Now \\( directmap^{\\prime}(steadfast) \\geqq 1 \\) implies\n\\[\nsmallness \\geqq \\int_{0}^{voidness} frozenpoint directmap^{\\prime}(frozenpoint)\\left[\\int_{0}^{frozenpoint} 2 steadfast d steadfast\\right] d frozenpoint=\\int_{0}^{voidness} frozenpoint^{3} directmap^{\\prime}(frozenpoint) d frozenpoint=littleness\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "y": "vbmncrst",
        "z": "kpltwhfq",
        "f": "wcrlxdge",
        "g": "ndfsopak",
        "G": "bqtrmzxe",
        "F": "lhgwpvso",
        "K": "sdjkrbmn",
        "c": "vghsneql",
        "A": "trqmdplk",
        "B": "szpcnwhm"
      },
      "question": "B-4. (a) On [0,1], let \\( wcrlxdge \\) have a continuous derivative satisfying \\( 0<wcrlxdge^{\\prime}(qzxwvtnp) \\leqq 1 \\). Also suppose that \\( wcrlxdge(0)=0 \\). Prove that\n\\[\n\\left[\\int_{0}^{1} wcrlxdge(qzxwvtnp) d qzxwvtnp\\right]^{2} \\geqq \\int_{0}^{1}[wcrlxdge(qzxwvtnp)]^{3} d qzxwvtnp\n\\]\n[Hint: Replace the inequality by one involving the inverse function to \\( wcrlxdge \\).]\n(b) Show an example in which equality occurs.",
      "solution": "B-4. We give two solutions; the first does not use the hint and the second does.\nTheorem. If \\( wcrlxdge \\) is continuous on \\( [0,1], wcrlxdge(0)=0 \\), and \\( 0 \\leqq wcrlxdge^{\\prime}(qzxwvtnp) \\leqq 1 \\) on \\( (0,1) \\), then\n\\[\n\\left[\\int_{0}^{1} wcrlxdge(qzxwvtnp) d qzxwvtnp\\right]^{2}>\\int_{0}^{1}[wcrlxdge(qzxwvtnp)]^{3} d qzxwvtnp\n\\]\nunless, identically on \\( [0,1] \\), either \\( wcrlxdge(qzxwvtnp)=qzxwvtnp \\) or \\( wcrlxdge(qzxwvtnp)=0 \\).\nProof. Define \\( bqtrmzxe(hjgrksla)=2 \\int_{0}^{hjgrksla} wcrlxdge(qzxwvtnp) d qzxwvtnp-[wcrlxdge(hjgrksla)]^{2} \\) for \\( hjgrksla \\in[0,1] \\). Then \\( bqtrmzxe(0)=0 \\) and \\( bqtrmzxe^{\\prime}(hjgrksla)=2 wcrlxdge(hjgrksla)\\left[1-wcrlxdge^{\\prime}(hjgrksla)\\right] \\geqq 0 \\), so that \\( bqtrmzxe(hjgrksla) \\geqq 0 \\) and consequently \\( wcrlxdge(hjgrksla) bqtrmzxe(hjgrksla) \\geqq 0 \\).\n\nNow define \\( lhgwpvso(hjgrksla)=\\left[\\int_{0}^{hjgrksla} wcrlxdge(qzxwvtnp) d qzxwvtnp\\right]^{2}-\\int_{0}^{hjgrksla}[wcrlxdge(qzxwvtnp)]^{3} d qzxwvtnp \\) for \\( hjgrksla \\in[0,1] \\). Then \\( lhgwpvso(0)=0 \\) and \\( lhgwpvso^{\\prime}(hjgrksla)=wcrlxdge(hjgrksla) bqtrmzxe(hjgrksla) \\geqq 0 \\), so that \\( lhgwpvso(hjgrksla) \\geqq 0 \\) and in particular \\( lhgwpvso(1) \\geqq 0 \\).\n\nEquality is possible only if \\( wcrlxdge(hjgrksla) bqtrmzxe(hjgrksla)=lhgwpvso^{\\prime}(hjgrksla)=0 \\) for all \\( hjgrksla \\), which implies that, for some \\( sdjkrbmn, wcrlxdge=0 \\) on \\( [0, sdjkrbmn] \\) and \\( bqtrmzxe^{\\prime}=0 \\), with \\( wcrlxdge>0 \\), on \\( (sdjkrbmn, 1) \\). We then have \\( wcrlxdge^{\\prime}=1 \\) on \\( (sdjkrbmn, 1) \\), which is admissible only if \\( sdjkrbmn=0 \\) or \\( sdjkrbmn=1 \\), since otherwise \\( wcrlxdge^{\\prime}(sdjkrbmn) \\) is simultaneously defined and undefined.\n\nThe unique answer to (b) is \\( wcrlxdge(qzxwvtnp)=qzxwvtnp \\). The following is an outline of a proof of (a) using the hint. Let \\( wcrlxdge(1)=vghsneql \\). The hypothesis implies that \\( wcrlxdge \\) has an inverse \\( ndfsopak \\) with \\( ndfsopak^{\\prime}(vbmncrst) \\geqq 1 \\) on \\( 0 \\leqq vbmncrst \\leqq vghsneql \\). Let\n\\[\ntrqmdplk=\\left[\\int_{0}^{1} wcrlxdge(qzxwvtnp) d qzxwvtnp\\right]^{2} \\text { and } szpcnwhm=\\int_{0}^{1}[wcrlxdge(qzxwvtnp)]^{3} d qzxwvtnp\n\\]\n\nThen\n\\[\ntrqmdplk=\\left[\\int_{0}^{vghsneql} vbmncrst \\, ndfsopak^{\\prime}(vbmncrst) d vbmncrst\\right]^{2}=\\int_{0}^{vghsneql} \\int_{0}^{vghsneql} vbmncrst \\, ndfsopak^{\\prime}(vbmncrst) \\, kpltwhfq \\, ndfsopak^{\\prime}(kpltwhfq) d kpltwhfq d vbmncrst=2 \\int_{0}^{vghsneql} \\int_{0}^{kpltwhfq} vbmncrst \\, ndfsopak^{\\prime}(vbmncrst) \\, kpltwhfq \\, ndfsopak^{\\prime}(kpltwhfq) d vbmncrst d kpltwhfq\n\\]\nusing the symmetry of the integrand about the line \\( vbmncrst=kpltwhfq \\). Now \\( ndfsopak^{\\prime}(vbmncrst) \\geqq 1 \\) implies\n\\[\ntrqmdplk \\geqq \\int_{0}^{vghsneql} kpltwhfq \\, ndfsopak^{\\prime}(kpltwhfq)\\left[\\int_{0}^{kpltwhfq} 2 vbmncrst d vbmncrst\\right] d kpltwhfq=\\int_{0}^{vghsneql} kpltwhfq^{3} ndfsopak^{\\prime}(kpltwhfq) d kpltwhfq=szpcnwhm\n\\]"
    },
    "kernel_variant": {
      "question": "Let $M>0$ be fixed and let \n\\[\nf:[0,1]\\longrightarrow[0,\\infty)\n\\]\nbe a $C^{1}$–function such that  \n\n\\[\n\\text{(i)}\\qquad 0\\le f'(x)\\le M \\quad (0<x<1),\\qquad\\qquad\n\\text{(ii)}\\qquad f(0)=0 .\n\\]\n\n(a)\\; Prove the sharp inequality  \n\\[\n\\Bigl(\\,\\int_{0}^{1} f(x)\\,dx\\Bigr)^{2}\\;\\ge\\;\n\\frac{1}{M}\\int_{0}^{1} f(x)^{\\,3}\\,dx.\n\\tag{$\\ast$}\n\\]\n\n(b)\\; Show that the constant $1/M$ in $(\\ast)$ cannot be replaced by any larger\npositive number that would still make the inequality valid for every\nfunction satisfying $(\\text{i})$–$(\\text{ii})$.\n\n(c)\\; Determine all functions $f$ for which equality occurs in $(\\ast)$.\n\n(d)\\; Establish the first–order (“limit $p\\to1^{+}$’’) variant of $(\\ast)$:\nprove that  \n\\[\n\\int_{0}^{1} f(x)\\,dx\\;\\ge\\;\n\\frac{3}{2M}\\int_{0}^{1} f(x)^{\\,2}\\,dx,\n\\tag{$\\dagger$}\n\\]\nand show that the constant $\\dfrac{3}{2M}$ is again best possible.",
      "solution": "Throughout we normalise by setting  \n\\[\ng(x)=\\frac{f(x)}{M}\\quad(0\\le x\\le1),\n\\qquad\\Longrightarrow\\qquad\ng(0)=0,\\qquad 0\\le g'(x)\\le 1 .\n\\]\nStatements $(\\ast),(\\dagger)$ are therefore equivalent to  \n\n\\[\n\\Bigl(\\int_{0}^{1}g\\Bigr)^{2}\\ge\\int_{0}^{1}g^{3},\n\\tag{$\\ast_{1}$}\n\\qquad\\qquad\n\\int_{0}^{1} g\\;\\ge\\;\\frac{3}{2}\\int_{0}^{1}g^{2}.\n\\tag{$\\dagger_{1}$}\n\\]\n\nPut  \n\\[\nF(t)=\\int_{0}^{t}g(x)\\,dx \\quad(0\\le t\\le1),\\qquad \nG(t)=2F(t)-g(t)^{2}.\n\\]\n\n--------------------------------------------------------------------\nStep 1.  A basic lemma (common to all parts).  \n\nBecause $0\\le g'(t)\\le1$ and $g(t)\\ge0$ one has  \n\n\\[\nG'(t)=2g(t)\\bigl[1-g'(t)\\bigr]\\ge0,\n\\]\nso that $G$ is increasing and $G(t)\\ge0$ on $[0,1]$.\n\n--------------------------------------------------------------------\nStep 2.  Proof of $(\\ast_{1})$ (part (a)).  \n\nDefine $H(t)=F(t)^{2}-\\int_{0}^{t} g(x)^{3}\\,dx$.  \nSince $F'(t)=g(t)$,\n\\[\nH'(t)=2F(t)g(t)-g(t)^{3}=g(t)\\,G(t)\\ge0,\n\\]\nwhence $H(1)\\ge H(0)=0$, i.e.\\ $(\\ast_{1})$; scaling back gives $(\\ast)$.\n\n--------------------------------------------------------------------\nStep 3.  Optimality of $1/M$ (part (b)).  \n\nFor $k\\in(0,M]$ let $f_{k}(x)=k x$.  \nConditions $(\\text{i})$–$(\\text{ii})$ hold and\n\\[\n\\frac{\\bigl(\\int_{0}^{1}f_{k}\\bigr)^{2}}{\\int_{0}^{1}f_{k}^{3}}\n=\\frac{1}{k}\\xrightarrow[k\\uparrow M]{}\\frac{1}{M},\n\\]\nso the constant $1/M$ is best possible.\n\n--------------------------------------------------------------------\nStep 4.  Equality in $(\\ast)$ (part (c)).  \n\nEquality in $(\\ast_{1})$ forces $H'(t)\\equiv0$, hence\n$g(t)G(t)\\equiv0$.  Because $g\\not\\equiv0$ we must have $G(t)\\equiv0$,\nthat is $2F(t)=g(t)^{2}$ for all $t$.  \nDifferentiation gives\n$2g(t)=2g(t)g'(t)$, so $g'(t)\\equiv1$ on every component where\n$g(t)\\neq0$.  \nContinuity of $g'$ leaves the single possibility $g(x)=x$;\ntherefore $f\\equiv0$ or $f(x)=Mx$.\n\n--------------------------------------------------------------------\nStep 5.  The first–order inequality $(\\dagger_{1})$ (part (d)).  \n\nThe proof is supplied in three rigorous sub–steps.\n\n--------------------------------------------------------------------\nStep 5.1 Reduction to an optimisation problem.  \n\nIntroduce\n\\[\nA:=\\int_{0}^{1}g(x)\\,dx,\\qquad\nB:=\\int_{0}^{1}g(x)^{2}\\,dx .\n\\]\nInequality $(\\dagger_{1})$ is $A\\ge\\tfrac{3}{2}B$, i.e.\\\n\\[\n\\frac{B}{A}\\le\\frac{2}{3}.\n\\tag{1}\n\\]\nFix $A>0$ and maximise $B$ under the constraints  \n\n\\[\n\\mathcal C:=\\bigl\\{g\\in C^{1}[0,1]\\;:\\;g(0)=0,\\;0\\le g'(x)\\le1,\\;\n\\int_{0}^{1} g =A \\bigr\\}.\n\\]\n\n--------------------------------------------------------------------\nStep 5.2 Convex‐analytic “bang–bang’’ principle.  \n\nLet $s(x)=g'(x)\\in[0,1]$ for $g\\in\\mathcal C$; then  \n$g(x)=\\int_{0}^{x}s(t)\\,dt$.\nIn terms of $s$ one has\n\\[\nA=\\int_{0}^{1}(1-x)\\,s(x)\\,dx ,\n\\qquad\nB=\\int_{0}^{1}\\Bigl(\\int_{0}^{x}s(t)\\,dt\\Bigr)^{2}dx .\n\\]\nThe admissible set  \n\\(\n\\mathcal S:=\\bigl\\{s\\in L^{\\infty}(0,1):0\\le s\\le1,\\,\n\\int_{0}^{1}(1-x)s(x)=A\\bigr\\}\n\\)\nis a weak* compact, convex subset of $L^{\\infty}(0,1)$.  \nThe functional $B$ is continuous and strictly convex in $s$ (a routine\nquadratic–form computation), therefore\nevery maximiser of $B$ over $\\mathcal S$ is an extreme point of\n$\\mathcal S$.  Kreĭn–Milman implies that such an extreme point is\nnecessarily of “bang–bang’’ type, i.e.\\ $s(x)\\in\\{0,1\\}$ a.e.  \n\nThus each maximiser is the derivative of a function\n\\[\ng_{c}(x)=\\max\\{0,x-c\\}\\qquad(0\\le c<1),\n\\tag{2}\n\\]\nobtained by staying flat on $[0,c]$ and rising with unit slope on\n$[c,1]$.  (Any $g_{c}$ can be approximated in $C^{1}$ by admissible\nfunctions, so optimising over the larger Lipschitz class does not\nchange the supremum.)\n\n--------------------------------------------------------------------\nStep 5.3 Computation on the extremal family.  \n\nWrite $d:=1-c=g_{c}(1)$.  Elementary integration gives  \n\\[\nA=\\int_{0}^{1}g_{c}\n      =\\frac{d^{2}}{2},\\qquad\nB=\\int_{0}^{1}g_{c}^{2}\n      =\\frac{d^{3}}{3},\n\\]\nhence  \n\\[\n\\frac{B}{A}=\\frac{\\tfrac{d^{3}}{3}}{\\tfrac{d^{2}}{2}}\n           =\\frac{2}{3}\\,d\\le\\frac{2}{3},\n\\]\nwith equality iff $d=1$, that is, $c=0$ and $g_{c}(x)=x$.\n\n--------------------------------------------------------------------\nStep 5.4 Conclusion.  \n\nFor every admissible $g$ we have $\\dfrac{B}{A}\\le\\dfrac23$,\nequivalently $A\\ge\\dfrac32 B$, proving $(\\dagger_{1})$.  \nThe extremisers are again $g\\equiv0$ and $g(x)=x$; rescaling yields\n$(\\dagger)$ for $f$ and shows the constant $\\dfrac{3}{2M}$ to be optimal.\n\n--------------------------------------------------------------------\nAll statements in parts (a)–(d) are now established rigorously, and\nevery constant has been proved sharp.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.610579",
        "was_fixed": false,
        "difficulty_analysis": "1. Variable exponent p and two–sided derivative bounds (m,M) force the solver\n   to keep track of scaling behaviour and to optimise the constant with\n   respect to all admissible slopes; this is far more delicate than the fixed\n   quadratic–cubic comparison in the original problem.\n\n2. The proof demands simultaneous use of:  \n   • monotonicity of increasing functions,  \n   • carefully chosen auxiliary functions H(t) and differential inequalities,  \n   • optimisation over a continuum of affine test functions to establish\n     sharpness, and  \n   • a limiting argument (p→1) involving logarithmic derivatives.\n\n3. The equality analysis shows that every inequality sign used must be rigid;\n   tracing this back through the chain of estimates requires a precise\n   understanding of when all intermediate inequalities become equalities.\n\n4. Altogether, the enhanced variant combines elementary integral estimates,\n   monotone function theory, optimisation, and limiting processes—going well\n   beyond the single-exponent, one–sided Lipschitz setting of the original\n   problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $M>0$ be fixed and let  \n  $f:[0,1]\\longrightarrow[0,\\infty)$  \nbe a $C^{1}$–function satisfying  \n\n  (i) $0\\le f'(x)\\le M\\qquad(0<x<1),$  \n\n  (ii) $f(0)=0.$  \n\n(a)  Prove the sharp inequality  \n                           \n                 $\\displaystyle\\Bigl(\\int_{0}^{1}f(x)\\,dx\\Bigr)^2\\;\\ge\\;\n                             \\frac1M\\int_{0}^{1}f(x)^{\\,3}\\,dx.$          $(\\ast)$  \n\n(b)  Show that the constant $1/M$ in $(\\ast)$ cannot be replaced by any larger\npositive number that would still make the inequality valid for all functions\nsatisfying (i)–(ii).\n\n(c)  Determine all functions $f$ for which equality holds in $(\\ast)$.\n\n(d)  Establish the first–order (“limit $p\\to1^{+}$”) variant of $(\\ast)$: prove that  \n                                       \n               $\\displaystyle\\int_{0}^{1}f(x)\\,dx\\;\\ge\\;\n                         \\frac{3}{2M}\\int_{0}^{1}f(x)^{\\,2}\\,dx,$  \n\nand show that the constant $3/(2M)$ is again best possible.",
      "solution": "Step 0.  Normalisation.  \nPut $g(x)=f(x)/M$.  Then $g(0)=0$ and $0\\le g'(x)\\le 1$.  \nInequality $(\\ast)$ is equivalent to  \n\n  $\\bigl(\\int_{0}^{1}g\\bigr)^{2}\\ge\\int_{0}^{1}g^{3},\\qquad$$(\\ast_{1})$\n\nso we work with $g$ and reinstate the factor $M$ only at the very end.\n\nThroughout the proof denote  \n  $F(t)=\\displaystyle\\int_{0}^{t}g(x)\\,dx\\qquad(0\\le t\\le1).$\n\n--------------------------------------------------------------------\nStep 1.  A basic non-negativity lemma.  \nDefine  \n\n  $G(t)=2\\int_{0}^{t}g(x)\\,dx-\\;g(t)^{2}=2F(t)-g(t)^{2}\\quad(0\\le t\\le1).$\n\nBecause $g'(t)\\le 1$ and $g(t)\\ge0$,  \n\n  $G'(t)=2g(t)\\bigl[1-g'(t)\\bigr]\\;\\ge\\;0,$  \n\nso $G$ is increasing and $G(t)\\ge 0$ for every $t$.\n\n--------------------------------------------------------------------\nStep 2.  Proof of $(\\ast_{1})$.  \nSet  \n\n  $H(t)=F(t)^{2}-\\int_{0}^{t}g(x)^{3}\\,dx\\quad(0\\le t\\le1).$\n\nBecause $F'(t)=g(t)$ we obtain  \n\n  $H'(t)=2F(t)g(t)-g(t)^{3}\n          =g(t)\\bigl[\\,2F(t)-g(t)^{2}\\bigr]\n          =g(t)\\,G(t)\\;\\ge\\;0.$\n\nHence $H$ is non-decreasing and  \n\n  $H(1)\\;\\ge\\;H(0)=0\\;,$ i.e. $\\displaystyle F(1)^{2}\\ge\\int_{0}^{1}g^{3},$\nwhich is $(\\ast_{1})$.  Restoring $f$ gives $(\\ast)$.\n\n--------------------------------------------------------------------\nStep 3.  Optimality of the constant (part (b)).  \nFor $k\\in[0,M]$ consider the test functions $f_{k}(x)=k\\,x$.  \nThey satisfy $f_{k}(0)=0$ and $f_{k}'(x)=k\\le M$.  Moreover  \n\n\\[\n\\int_{0}^{1}f_{k}= \\frac{k}{2},\\qquad\n\\int_{0}^{1}f_{k}^{3}=k^{3}\\!\\int_{0}^{1}x^{3}dx=\\frac{k^{3}}{4},\n\\]\n\nso  \n\n\\[\n\\frac{\\bigl(\\int_{0}^{1}f_{k}\\bigr)^{2}}{\\int_{0}^{1}f_{k}^{3}}\n     =\\frac{k^{2}/4}{k^{3}/4}= \\frac1k.\n\\]\n\nBecause $k\\uparrow M$ makes this ratio approach $1/M$, no larger\nconstant than $1/M$ can work in $(\\ast)$, proving sharpness.\n\n--------------------------------------------------------------------\nStep 4.  Equality case (part (c)).  \nEquality in $(\\ast_{1})$ forces $H'(t)\\equiv0$, i.e. $g(t)\\,G(t)\\equiv0$.\nSince $g\\not\\equiv0$ (otherwise $(\\ast)$ is trivial) we must have\n$G(t)\\equiv0$, meaning $2F(t)=g(t)^{2}$ for all $t$.  Differentiating\ngives $2g(t)=2g(t)g'(t)$, hence $g'(t)\\equiv1$ wherever $g(t)\\neq0$.\nBecause $g$ and $g'$ are continuous, the only possibility is\n\n  $g(x)=x\\quad(0\\le x\\le1).$\n\nRetranslating, the equality cases are  \n\n  $f\\equiv0\\quad\\text{and}\\quad f(x)=Mx\\;(0\\le x\\le1).$\n\n--------------------------------------------------------------------\nStep 5.  The first–order variant (part (d)).  \nDefine the auxiliary functional  \n\n  $Q(t)=3F(t)-2\\int_{0}^{t}g(x)^{2}\\,dx\\qquad(0\\le t\\le1).$\n\nBecause $F'(t)=g(t)$ we have  \n\n  $Q'(t)=3g(t)-2g(t)^{2}=g(t)\\bigl[\\,3-2g(t)\\bigr].$\n\nNow $g'(x)\\le1$ and $g(0)=0$ imply $g(x)=\\int_{0}^{x}g'(s)\\,ds\\le x$,\nhence $0\\le g(t)\\le1$ for every $t$.  Consequently $3-2g(t)\\ge1>0$,\nand together with $g(t)\\ge0$ we obtain $Q'(t)\\ge0$.  Thus $Q$ is\nnon-decreasing and  \n\n\\[\n3F(1)-2\\int_{0}^{1}g^{2}\\ge Q(1)\\ge Q(0)=0,\n\\]\ni.e.  \n\n\\[\n\\int_{0}^{1}g(x)\\,dx\\;\\ge\\;\\frac32\\int_{0}^{1}g(x)^{2}\\,dx.\n\\]\n\nReturning to $f$ yields  \n\n\\[\n\\int_{0}^{1}f(x)\\,dx\\;\\ge\\;\\frac{3}{2M}\\int_{0}^{1}f(x)^{2}\\,dx.\n\\]\n\nSharpness.  \nFor the linear extremal $f(x)=Mx$ we have  \n\n\\[\n\\int_{0}^{1}f=\\frac{M}{2},\\qquad\n\\int_{0}^{1}f^{2}=M^{2}\\!\\int_{0}^{1}x^{2}dx=\\frac{M^{2}}{3},\n\\]\nand indeed  \n\\[\n\\frac{\\int_{0}^{1}f}{\\int_{0}^{1}f^{2}}=\\frac{3}{2M},\n\\]\nso the constant $3/(2M)$ cannot be improved.\n\n--------------------------------------------------------------------\nEverything asked for in parts (a)–(d) is now established rigorously.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.488948",
        "was_fixed": false,
        "difficulty_analysis": "1. Variable exponent p and two–sided derivative bounds (m,M) force the solver\n   to keep track of scaling behaviour and to optimise the constant with\n   respect to all admissible slopes; this is far more delicate than the fixed\n   quadratic–cubic comparison in the original problem.\n\n2. The proof demands simultaneous use of:  \n   • monotonicity of increasing functions,  \n   • carefully chosen auxiliary functions H(t) and differential inequalities,  \n   • optimisation over a continuum of affine test functions to establish\n     sharpness, and  \n   • a limiting argument (p→1) involving logarithmic derivatives.\n\n3. The equality analysis shows that every inequality sign used must be rigid;\n   tracing this back through the chain of estimates requires a precise\n   understanding of when all intermediate inequalities become equalities.\n\n4. Altogether, the enhanced variant combines elementary integral estimates,\n   monotone function theory, optimisation, and limiting processes—going well\n   beyond the single-exponent, one–sided Lipschitz setting of the original\n   problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}