{
  "index": "1967-B-6",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-6. Let \\( f \\) beareal-valued function having partial derivatives and which is defined for \\( x^{2}+y^{2} \\leq 1 \\) and is such that \\( |f(x, y)| \\leqq 1 \\). Show that there exists a point ( \\( x_{0}, y_{0} \\) ) in the interior of the unit circle such that\n\\[\n\\left(\\frac{\\partial f}{\\partial x}\\left(x_{0}, y_{0}\\right)\\right)^{2}+\\left(\\frac{\\partial f}{\\partial y}\\left(x_{0}, y_{0}\\right)\\right)^{2} \\leqq 16\n\\]",
  "solution": "B-6 Consider the function \\( g \\) whose values are defined by \\( g(x, y)=f(x, y) \\) \\( +2\\left(x^{2}+y^{2}\\right) \\). On the circumference of the unit circle, \\( g(x, y) \\geqq 1 \\), and at the origin, \\( g(0,0) \\leqq 1 \\). Hence, either \\( g(x, y)= \\) constant and \\( f(x, y)= \\) constant \\( -2\\left(x^{2}+y^{2}\\right) \\), or there is a minimum value for \\( g(x, y) \\) at some interior point. In the case \\( g(x, y)= \\) constant, the result is immediate. Otherwise, let \\( \\left(x_{0}, y_{0}\\right) \\) be the coordinates of a point where \\( g(x, y) \\) has a minimum. Then\n\\[\n\\begin{array}{c}\n\\frac{\\partial g}{\\partial x}=\\frac{\\partial g}{\\partial y}=0 \\text { at }\\left(x_{0}, y_{0}\\right) \\\\\n\\text { and }\\left|\\frac{\\partial f}{\\partial x}\\left(x_{0}, y_{0}\\right)\\right| \\leqq 4\\left|x_{0}\\right|,\\left|\\frac{\\partial f}{\\partial y}\\left(x_{0}, y_{0}\\right)\\right| \\leqq 4\\left|y_{0}\\right|\n\\end{array}\n\\]\n\nThus the conclusion follows.",
  "vars": [
    "f",
    "g",
    "x",
    "y",
    "x_0",
    "y_0"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "realfunc",
        "g": "auxifun",
        "x": "axisxcoor",
        "y": "axisycoor",
        "x_0": "focusxpt",
        "y_0": "focusypt"
      },
      "question": "B-6. Let \\( realfunc \\) be a real-valued function having partial derivatives and which is defined for \\( axisxcoor^{2}+axisycoor^{2} \\leq 1 \\) and is such that \\( |realfunc(axisxcoor, axisycoor)| \\leqq 1 \\). Show that there exists a point ( \\( focusxpt, focusypt \\) ) in the interior of the unit circle such that\n\\[\n\\left(\\frac{\\partial realfunc}{\\partial axisxcoor}\\left(focusxpt, focusypt\\right)\\right)^{2}+\\left(\\frac{\\partial realfunc}{\\partial axisycoor}\\left(focusxpt, focusypt\\right)\\right)^{2} \\leqq 16\n\\]",
      "solution": "B-6 Consider the function \\( auxifun \\) whose values are defined by \\( auxifun(axisxcoor, axisycoor)=realfunc(axisxcoor, axisycoor)+2\\left(axisxcoor^{2}+axisycoor^{2}\\right) \\). On the circumference of the unit circle, \\( auxifun(axisxcoor, axisycoor) \\geqq 1 \\), and at the origin, \\( auxifun(0,0) \\leqq 1 \\). Hence, either \\( auxifun(axisxcoor, axisycoor)= \\) constant and \\( realfunc(axisxcoor, axisycoor)= \\) constant \\( -2\\left(axisxcoor^{2}+axisycoor^{2}\\right) \\), or there is a minimum value for \\( auxifun(axisxcoor, axisycoor) \\) at some interior point. In the case \\( auxifun(axisxcoor, axisycoor)= \\) constant, the result is immediate. Otherwise, let \\( \\left(focusxpt, focusypt\\right) \\) be the coordinates of a point where \\( auxifun(axisxcoor, axisycoor) \\) has a minimum. Then\n\\[\n\\begin{array}{c}\n\\frac{\\partial auxifun}{\\partial axisxcoor}=\\frac{\\partial auxifun}{\\partial axisycoor}=0 \\text { at }\\left(focusxpt, focusypt\\right) \\\\\n\\text { and }\\left|\\frac{\\partial realfunc}{\\partial axisxcoor}\\left(focusxpt, focusypt\\right)\\right| \\leqq 4\\left|focusxpt\\right|,\\left|\\frac{\\partial realfunc}{\\partial axisycoor}\\left(focusxpt, focusypt\\right)\\right| \\leqq 4\\left|focusypt\\right|\n\\end{array}\n\\]\n\nThus the conclusion follows."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "sandstone",
        "g": "gearshift",
        "x": "pendulum",
        "y": "telescope",
        "x_0": "pendulumroot",
        "y_0": "telescopezero"
      },
      "question": "B-6. Let \\( sandstone \\) beareal-valued function having partial derivatives and which is defined for \\( pendulum^{2}+telescope^{2} \\leq 1 \\) and is such that \\( |sandstone(pendulum, telescope)| \\leqq 1 \\). Show that there exists a point ( \\( pendulumroot, telescopezero \\) ) in the interior of the unit circle such that\n\\[\n\\left(\\frac{\\partial sandstone}{\\partial pendulum}\\left(pendulumroot, telescopezero\\right)\\right)^{2}+\\left(\\frac{\\partial sandstone}{\\partial telescope}\\left(pendulumroot, telescopezero\\right)\\right)^{2} \\leqq 16\n\\]",
      "solution": "B-6 Consider the function \\( gearshift \\) whose values are defined by \\( gearshift(pendulum, telescope)=sandstone(pendulum, telescope) \\) \\( +2\\left(pendulum^{2}+telescope^{2}\\right) \\). On the circumference of the unit circle, \\( gearshift(pendulum, telescope) \\geqq 1 \\), and at the origin, \\( gearshift(0,0) \\leqq 1 \\). Hence, either \\( gearshift(pendulum, telescope)= \\) constant and \\( sandstone(pendulum, telescope)= \\) constant \\( -2\\left(pendulum^{2}+telescope^{2}\\right) \\), or there is a minimum value for \\( gearshift(pendulum, telescope) \\) at some interior point. In the case \\( gearshift(pendulum, telescope)= \\) constant, the result is immediate. Otherwise, let \\( \\left(pendulumroot, telescopezero\\right) \\) be the coordinates of a point where \\( gearshift(pendulum, telescope) \\) has a minimum. Then\n\\[\n\\begin{array}{c}\n\\frac{\\partial gearshift}{\\partial pendulum}=\\frac{\\partial gearshift}{\\partial telescope}=0 \\text { at }\\left(pendulumroot, telescopezero\\right) \\\\\n\\text { and }\\left|\\frac{\\partial sandstone}{\\partial pendulum}\\left(pendulumroot, telescopezero\\right)\\right| \\leqq 4\\left|pendulumroot\\right|,\\left|\\frac{\\partial sandstone}{\\partial telescope}\\left(pendulumroot, telescopezero\\right)\\right| \\leqq 4\\left|telescopezero\\right|\n\\end{array}\n\\]\n\nThus the conclusion follows."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "constantvalue",
        "g": "reducedfunction",
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "x_0": "variablevertical",
        "y_0": "variablehorizontal"
      },
      "question": "B-6. Let \\( constantvalue \\) beareal-valued function having partial derivatives and which is defined for \\( verticalaxis^{2}+horizontalaxis^{2} \\leq 1 \\) and is such that \\( |constantvalue(verticalaxis, horizontalaxis)| \\leqq 1 \\). Show that there exists a point ( \\( variablevertical, variablehorizontal \\) ) in the interior of the unit circle such that\n\\[\n\\left(\\frac{\\partial constantvalue}{\\partial verticalaxis}\\left(variablevertical, variablehorizontal\\right)\\right)^{2}+\\left(\\frac{\\partial constantvalue}{\\partial horizontalaxis}\\left(variablevertical, variablehorizontal\\right)\\right)^{2} \\leqq 16\n\\]\n",
      "solution": "B-6 Consider the function \\( reducedfunction \\) whose values are defined by \\( reducedfunction(verticalaxis, horizontalaxis)=constantvalue(verticalaxis, horizontalaxis) \\) \\( +2\\left(verticalaxis^{2}+horizontalaxis^{2}\\right) \\). On the circumference of the unit circle, \\( reducedfunction(verticalaxis, horizontalaxis) \\geqq 1 \\), and at the origin, \\( reducedfunction(0,0) \\leqq 1 \\). Hence, either \\( reducedfunction(verticalaxis, horizontalaxis)= \\) constant and \\( constantvalue(verticalaxis, horizontalaxis)= \\) constant \\( -2\\left(verticalaxis^{2}+horizontalaxis^{2}\\right) \\), or there is a minimum value for \\( reducedfunction(verticalaxis, horizontalaxis) \\) at some interior point. In the case \\( reducedfunction(verticalaxis, horizontalaxis)= \\) constant, the result is immediate. Otherwise, let \\( \\left(variablevertical, variablehorizontal\\right) \\) be the coordinates of a point where \\( reducedfunction(verticalaxis, horizontalaxis) \\) has a minimum. Then\n\\[\n\\begin{array}{c}\n\\frac{\\partial reducedfunction}{\\partial verticalaxis}=\\frac{\\partial reducedfunction}{\\partial horizontalaxis}=0 \\text { at }\\left(variablevertical, variablehorizontal\\right) \\\\\n\\text { and }\\left|\\frac{\\partial constantvalue}{\\partial verticalaxis}\\left(variablevertical, variablehorizontal\\right)\\right| \\leqq 4\\left|variablevertical\\right|,\\left|\\frac{\\partial constantvalue}{\\partial horizontalaxis}\\left(variablevertical, variablehorizontal\\right)\\right| \\leqq 4\\left|variablehorizontal\\right|\n\\end{array}\n\\]\n\nThus the conclusion follows."
    },
    "garbled_string": {
      "map": {
        "f": "ksqnvlaj",
        "g": "zpdtmwhk",
        "x": "qxbbncyt",
        "y": "rvmhjlsa",
        "x_0": "qxklpnvr",
        "y_0": "rjdfmnes"
      },
      "question": "B-6. Let \\( ksqnvlaj \\) be a real-valued function having partial derivatives and which is defined for \\( qxbbncyt^{2}+rvmhjlsa^{2} \\leq 1 \\) and is such that \\( |ksqnvlaj(qxbbncyt, rvmhjlsa)| \\leqq 1 \\). Show that there exists a point ( \\( qxklpnvr, rjdfmnes \\) ) in the interior of the unit circle such that\n\\[\n\\left(\\frac{\\partial ksqnvlaj}{\\partial qxbbncyt}\\left(qxklpnvr, rjdfmnes\\right)\\right)^{2}+\\left(\\frac{\\partial ksqnvlaj}{\\partial rvmhjlsa}\\left(qxklpnvr, rjdfmnes\\right)\\right)^{2} \\leqq 16\n\\]",
      "solution": "B-6 Consider the function \\( zpdtmwhk \\) whose values are defined by \\( zpdtmwhk(qxbbncyt, rvmhjlsa)=ksqnvlaj(qxbbncyt, rvmhjlsa) +2\\left(qxbbncyt^{2}+rvmhjlsa^{2}\\right) \\). On the circumference of the unit circle, \\( zpdtmwhk(qxbbncyt, rvmhjlsa) \\geqq 1 \\), and at the origin, \\( zpdtmwhk(0,0) \\leqq 1 \\). Hence, either \\( zpdtmwhk(qxbbncyt, rvmhjlsa)= \\) constant and \\( ksqnvlaj(qxbbncyt, rvmhjlsa)= \\) constant \\( -2\\left(qxbbncyt^{2}+rvmhjlsa^{2}\\right) \\), or there is a minimum value for \\( zpdtmwhk(qxbbncyt, rvmhjlsa) \\) at some interior point. In the case \\( zpdtmwhk(qxbbncyt, rvmhjlsa)= \\) constant, the result is immediate. Otherwise, let \\( \\left(qxklpnvr, rjdfmnes\\right) \\) be the coordinates of a point where \\( zpdtmwhk(qxbbncyt, rvmhjlsa) \\) has a minimum. Then\n\\[\n\\begin{array}{c}\n\\frac{\\partial zpdtmwhk}{\\partial qxbbncyt}=\\frac{\\partial zpdtmwhk}{\\partial rvmhjlsa}=0 \\text { at }\\left(qxklpnvr, rjdfmnes\\right) \\\\\n\\text { and }\\left|\\frac{\\partial ksqnvlaj}{\\partial qxbbncyt}\\left(qxklpnvr, rjdfmnes\\right)\\right| \\leqq 4\\left|qxklpnvr\\right|,\\left|\\frac{\\partial ksqnvlaj}{\\partial rvmhjlsa}\\left(qxklpnvr, rjdfmnes\\right)\\right| \\leqq 4\\left|rjdfmnes\\right|\n\\end{array}\n\\]\n\nThus the conclusion follows."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ and let $A$ be a real, symmetric, positive–definite $n\\times n$ matrix.  \nIntroduce the quadratic form and the associated (closed and open) ellipsoids  \n\\[\nQ(x)=x^{\\mathsf T}Ax ,\\qquad \n\\Omega=\\{x\\in\\mathbb R^{n}:Q(x)\\le 1\\},\\qquad \n\\Omega^{\\circ}=\\{x\\in\\mathbb R^{n}:Q(x)<1\\}.\n\\]\n\nAssume a function $f:\\overline{\\Omega}\\longrightarrow\\mathbb R$ satisfies  \n\\[\nf\\in C^{1}(\\overline{\\Omega}),\\qquad |f(x)|\\le 1\\quad(\\forall\\,x\\in\\Omega).\n\\]\n\na)  Prove that there exists a point $x_{0}\\in\\Omega^{\\circ}$ such that  \n\\[\n\\bigl(\\nabla f(x_{0})\\bigr)^{\\mathsf T}A^{-1}\\,\\nabla f(x_{0})\\;<\\;16 .\n\\tag{1}\n\\]\n\nb)  Show that the constant $16$ in part (a) cannot be replaced by any {\\em smaller} universal\nnumber.  More precisely, prove that for every $\\varepsilon>0$ one can find a function\n$f_{\\varepsilon}\\in C^{1}(\\overline{\\Omega})$ with $|f_{\\varepsilon}(x)|\\le 1$\nand a point $x_{\\varepsilon}\\in\\Omega^{\\circ}$ for which  \n\\[\n\\bigl(\\nabla f_{\\varepsilon}(x_{\\varepsilon})\\bigr)^{\\mathsf T}A^{-1}\\,\n      \\nabla f_{\\varepsilon}(x_{\\varepsilon})\n      \\;>\\;16-\\varepsilon .\n\\tag{2}\n\\]\n\n(Hence the upper bound $16$ in (1) is best possible; it can be approached\nfrom below but never exceeded for the special point guaranteed in part (a).)",
      "solution": "Throughout we write $Q(x)=x^{\\mathsf T}Ax$; note $\\nabla Q(x)=2Ax$.\n\n--------------------------------------------------------------------\nPart (a).  Existence of an interior point with the strict bound (1).\n\nDefine\n\\[\ng(x)=f(x)+2\\,Q(x)\\quad (x\\in\\overline{\\Omega}).\n\\]\nBecause $g$ is continuous on the compact set $\\overline{\\Omega}$ it has\na global minimum, say at $x_{0}\\in\\overline{\\Omega}$.  \nOn the boundary $\\partial\\Omega$ we have $Q=1$ and hence\n\\[\ng(x)\\;=\\;f(x)+2\\;\\ge\\;-1+2\\;=\\;1\\qquad (x\\in\\partial\\Omega).\n\\tag{3}\n\\]\nAt the origin $g(0)=f(0)\\le1$.  Consequently $\\min_{\\overline{\\Omega}}g\\le1$,\nand relation (3) forces every minimiser to lie in $\\Omega^{\\circ}$.  Thus\n\\[\nx_{0}\\in\\Omega^{\\circ},\\qquad g(x_{0})=\\min_{\\overline{\\Omega}}g .\n\\tag{4}\n\\]\n\nSince $x_{0}$ is an {\\em interior} minimiser we have $\\nabla g(x_{0})=0$.\nBecause $\\nabla Q(x)=2Ax$, this means\n\\[\n\\nabla f(x_{0})=-4Ax_{0}.\n\\tag{5}\n\\]\nA short computation then gives\n\\[\n\\bigl(\\nabla f(x_{0})\\bigr)^{\\mathsf T}A^{-1}\\,\\nabla f(x_{0})\n      =16\\,x_{0}^{\\mathsf T}Ax_{0}\n      =16\\,Q(x_{0})\\;<\\;16 ,\n\\]\nbecause $x_{0}\\in\\Omega^{\\circ}$ implies $Q(x_{0})<1$.  Inequality (1)\nis proved.\n\n--------------------------------------------------------------------\nPart (b).  Sharpness of the constant.\n\nFix $\\varepsilon>0$ and put\n\\[\n\\delta=\\frac{\\varepsilon}{32}\\quad(0<\\delta<\\tfrac12).\n\\]\nIntroduce a smooth function $\\psi_{\\delta}:[0,1]\\to[-1,1]$ satisfying  \n\n(i) $\\psi_{\\delta}(t)=-1$ for $t\\in[1-\\delta,1]$;  \n\n(ii) $\\psi_{\\delta}(t)=1-2t$ for $t\\in[0,1-2\\delta]$;  \n\n(iii) $\\psi_{\\delta}$ is monotone decreasing and \n      $|\\psi_{\\delta}'(t)|\\le 4$ everywhere.\n\n(An explicit construction is easy: splice the linear\npiece $1-2t$ with the constant $-1$ by means of a standard mollifier\non the interval $[1-2\\delta,1-\\delta]$.)\n\nNow define\n\\[\nf_{\\varepsilon}(x)=\\psi_{\\delta}\\!\\bigl(Q(x)\\bigr),\\qquad x\\in\\overline{\\Omega}.\n\\]\nBecause $-1\\le\\psi_{\\delta}\\le 1$, the bound $|f_{\\varepsilon}|\\le1$\nis automatic.  We next locate a point where the gradient almost attains\nthe critical value $\\sqrt{16}$.\n\nChoose $t_{\\varepsilon}=1-2\\delta$ and select\n$x_{\\varepsilon}\\in\\Omega^{\\circ}$ with $Q(x_{\\varepsilon})=t_{\\varepsilon}$\n(arbitrarily, e.g.\\ on the major semi–axis of the ellipsoid).\nSince $\\psi_{\\delta}$ coincides with $1-2t$ in a neighbourhood of\n$t_{\\varepsilon}$, we have\n\\[\nf_{\\varepsilon}(x)=1-2Q(x)\\quad\\text{for }Q(x)\\le t_{\\varepsilon}+\\delta.\n\\]\nIn particular, $\\psi_{\\delta}'(t_{\\varepsilon})=-2$ and the chain rule yields\n\\[\n\\nabla f_{\\varepsilon}(x_{\\varepsilon})\n      =\\psi_{\\delta}'(t_{\\varepsilon})\\,\\nabla Q(x_{\\varepsilon})\n      =-2\\cdot 2Ax_{\\varepsilon}\n      =-4Ax_{\\varepsilon}.\n\\]\nConsequently\n\\[\n\\bigl(\\nabla f_{\\varepsilon}(x_{\\varepsilon})\\bigr)^{\\mathsf T}\n           A^{-1}\\,\\nabla f_{\\varepsilon}(x_{\\varepsilon})\n      =16\\,x_{\\varepsilon}^{\\mathsf T}Ax_{\\varepsilon}\n      =16\\,Q(x_{\\varepsilon})\n      =16\\,(1-2\\delta)\n      =16-\\varepsilon .\n\\]\nBecause $t_{\\varepsilon}<1$, the point $x_{\\varepsilon}$ is indeed\ninside $\\Omega$.  Replacing $\\varepsilon$ by $\\varepsilon/2$ if\nnecessary we obtain the {\\em strict} inequality (2).\n\nSince $\\varepsilon>0$ was arbitrary, the upper bound $16$ established in\npart (a) cannot be diminished: for any smaller constant $c<16$ one could\ntake $\\varepsilon=16-c$ and construct an $f_{\\varepsilon}$ that violates\nthe putative bound.  Thus $16$ is optimal.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.574179",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension and anisotropy:  The domain is an arbitrary ellipsoid in ℝⁿ, encoded by an n×n positive-definite matrix A, instead of a unit disc.  Competitors must manipulate quadratic forms and matrix norms, not merely Euclidean lengths.\n\n2.  Non-Euclidean metric in the conclusion:  The bound involves the A⁻¹-weighted norm ∇fᵀA⁻¹∇f, demanding familiarity with congruence of symmetric forms and spectral bounds (eigenvalues).\n\n3.  Optimal-constant proof:  Part (b) forces contestants to exhibit near-extremal examples, not just establish an inequality.  Producing such sharpness examples typically requires deeper insight into how the auxiliary function technique saturates the bound.\n\n4.  Multiple interacting ideas:  The solution mingles classical calculus of minima, matrix analysis (spectral radius estimates), and an extremal construction.  The original problem needed only a single auxiliary function; here one must devise both the minimising argument and an explicit asymptotically optimal family, adding a second layer of reasoning.\n\n5.  Technical precision:  Competitors must verify differentiability on an ellipsoid, justify why a minimum exists in Ω, manipulate gradients of quadratic forms in matrix notation, and keep track of operator inverses – all absent from the original two-dimensional, isotropic setting.\n\nHence the enhanced variant is markedly more intricate, demands broader mathematical tools, and carries a non-trivial optimality component, making it substantially harder than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 2$ and let $A$ be a real, symmetric, positive–definite $n\\times n$ matrix.  \nIntroduce the quadratic form and the associated (closed and open) ellipsoids  \n\\[\nQ(x)=x^{\\mathsf T}Ax ,\\qquad \n\\Omega=\\{x\\in\\mathbb R^{n}:Q(x)\\le 1\\},\\qquad \n\\Omega^{\\circ}=\\{x\\in\\mathbb R^{n}:Q(x)<1\\}.\n\\]\n\nAssume a function $f:\\overline{\\Omega}\\longrightarrow\\mathbb R$ satisfies  \n\\[\nf\\in C^{1}(\\overline{\\Omega}),\\qquad |f(x)|\\le 1\\quad(\\forall\\,x\\in\\Omega).\n\\]\n\na)  Prove that there exists a point $x_{0}\\in\\Omega^{\\circ}$ such that  \n\\[\n\\bigl(\\nabla f(x_{0})\\bigr)^{\\mathsf T}A^{-1}\\,\\nabla f(x_{0})\\;<\\;16 .\n\\tag{1}\n\\]\n\nb)  Show that the constant $16$ in part (a) cannot be replaced by any {\\em smaller} universal\nnumber.  More precisely, prove that for every $\\varepsilon>0$ one can find a function\n$f_{\\varepsilon}\\in C^{1}(\\overline{\\Omega})$ with $|f_{\\varepsilon}(x)|\\le 1$\nand a point $x_{\\varepsilon}\\in\\Omega^{\\circ}$ for which  \n\\[\n\\bigl(\\nabla f_{\\varepsilon}(x_{\\varepsilon})\\bigr)^{\\mathsf T}A^{-1}\\,\n      \\nabla f_{\\varepsilon}(x_{\\varepsilon})\n      \\;>\\;16-\\varepsilon .\n\\tag{2}\n\\]\n\n(Hence the upper bound $16$ in (1) is best possible; it can be approached\nfrom below but never exceeded for the special point guaranteed in part (a).)",
      "solution": "Throughout we write $Q(x)=x^{\\mathsf T}Ax$; note $\\nabla Q(x)=2Ax$.\n\n--------------------------------------------------------------------\nPart (a).  Existence of an interior point with the strict bound (1).\n\nDefine\n\\[\ng(x)=f(x)+2\\,Q(x)\\quad (x\\in\\overline{\\Omega}).\n\\]\nBecause $g$ is continuous on the compact set $\\overline{\\Omega}$ it has\na global minimum, say at $x_{0}\\in\\overline{\\Omega}$.  \nOn the boundary $\\partial\\Omega$ we have $Q=1$ and hence\n\\[\ng(x)\\;=\\;f(x)+2\\;\\ge\\;-1+2\\;=\\;1\\qquad (x\\in\\partial\\Omega).\n\\tag{3}\n\\]\nAt the origin $g(0)=f(0)\\le1$.  Consequently $\\min_{\\overline{\\Omega}}g\\le1$,\nand relation (3) forces every minimiser to lie in $\\Omega^{\\circ}$.  Thus\n\\[\nx_{0}\\in\\Omega^{\\circ},\\qquad g(x_{0})=\\min_{\\overline{\\Omega}}g .\n\\tag{4}\n\\]\n\nSince $x_{0}$ is an {\\em interior} minimiser we have $\\nabla g(x_{0})=0$.\nBecause $\\nabla Q(x)=2Ax$, this means\n\\[\n\\nabla f(x_{0})=-4Ax_{0}.\n\\tag{5}\n\\]\nA short computation then gives\n\\[\n\\bigl(\\nabla f(x_{0})\\bigr)^{\\mathsf T}A^{-1}\\,\\nabla f(x_{0})\n      =16\\,x_{0}^{\\mathsf T}Ax_{0}\n      =16\\,Q(x_{0})\\;<\\;16 ,\n\\]\nbecause $x_{0}\\in\\Omega^{\\circ}$ implies $Q(x_{0})<1$.  Inequality (1)\nis proved.\n\n--------------------------------------------------------------------\nPart (b).  Sharpness of the constant.\n\nFix $\\varepsilon>0$ and put\n\\[\n\\delta=\\frac{\\varepsilon}{32}\\quad(0<\\delta<\\tfrac12).\n\\]\nIntroduce a smooth function $\\psi_{\\delta}:[0,1]\\to[-1,1]$ satisfying  \n\n(i) $\\psi_{\\delta}(t)=-1$ for $t\\in[1-\\delta,1]$;  \n\n(ii) $\\psi_{\\delta}(t)=1-2t$ for $t\\in[0,1-2\\delta]$;  \n\n(iii) $\\psi_{\\delta}$ is monotone decreasing and \n      $|\\psi_{\\delta}'(t)|\\le 4$ everywhere.\n\n(An explicit construction is easy: splice the linear\npiece $1-2t$ with the constant $-1$ by means of a standard mollifier\non the interval $[1-2\\delta,1-\\delta]$.)\n\nNow define\n\\[\nf_{\\varepsilon}(x)=\\psi_{\\delta}\\!\\bigl(Q(x)\\bigr),\\qquad x\\in\\overline{\\Omega}.\n\\]\nBecause $-1\\le\\psi_{\\delta}\\le 1$, the bound $|f_{\\varepsilon}|\\le1$\nis automatic.  We next locate a point where the gradient almost attains\nthe critical value $\\sqrt{16}$.\n\nChoose $t_{\\varepsilon}=1-2\\delta$ and select\n$x_{\\varepsilon}\\in\\Omega^{\\circ}$ with $Q(x_{\\varepsilon})=t_{\\varepsilon}$\n(arbitrarily, e.g.\\ on the major semi–axis of the ellipsoid).\nSince $\\psi_{\\delta}$ coincides with $1-2t$ in a neighbourhood of\n$t_{\\varepsilon}$, we have\n\\[\nf_{\\varepsilon}(x)=1-2Q(x)\\quad\\text{for }Q(x)\\le t_{\\varepsilon}+\\delta.\n\\]\nIn particular, $\\psi_{\\delta}'(t_{\\varepsilon})=-2$ and the chain rule yields\n\\[\n\\nabla f_{\\varepsilon}(x_{\\varepsilon})\n      =\\psi_{\\delta}'(t_{\\varepsilon})\\,\\nabla Q(x_{\\varepsilon})\n      =-2\\cdot 2Ax_{\\varepsilon}\n      =-4Ax_{\\varepsilon}.\n\\]\nConsequently\n\\[\n\\bigl(\\nabla f_{\\varepsilon}(x_{\\varepsilon})\\bigr)^{\\mathsf T}\n           A^{-1}\\,\\nabla f_{\\varepsilon}(x_{\\varepsilon})\n      =16\\,x_{\\varepsilon}^{\\mathsf T}Ax_{\\varepsilon}\n      =16\\,Q(x_{\\varepsilon})\n      =16\\,(1-2\\delta)\n      =16-\\varepsilon .\n\\]\nBecause $t_{\\varepsilon}<1$, the point $x_{\\varepsilon}$ is indeed\ninside $\\Omega$.  Replacing $\\varepsilon$ by $\\varepsilon/2$ if\nnecessary we obtain the {\\em strict} inequality (2).\n\nSince $\\varepsilon>0$ was arbitrary, the upper bound $16$ established in\npart (a) cannot be diminished: for any smaller constant $c<16$ one could\ntake $\\varepsilon=16-c$ and construct an $f_{\\varepsilon}$ that violates\nthe putative bound.  Thus $16$ is optimal.\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.465948",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension and anisotropy:  The domain is an arbitrary ellipsoid in ℝⁿ, encoded by an n×n positive-definite matrix A, instead of a unit disc.  Competitors must manipulate quadratic forms and matrix norms, not merely Euclidean lengths.\n\n2.  Non-Euclidean metric in the conclusion:  The bound involves the A⁻¹-weighted norm ∇fᵀA⁻¹∇f, demanding familiarity with congruence of symmetric forms and spectral bounds (eigenvalues).\n\n3.  Optimal-constant proof:  Part (b) forces contestants to exhibit near-extremal examples, not just establish an inequality.  Producing such sharpness examples typically requires deeper insight into how the auxiliary function technique saturates the bound.\n\n4.  Multiple interacting ideas:  The solution mingles classical calculus of minima, matrix analysis (spectral radius estimates), and an extremal construction.  The original problem needed only a single auxiliary function; here one must devise both the minimising argument and an explicit asymptotically optimal family, adding a second layer of reasoning.\n\n5.  Technical precision:  Competitors must verify differentiability on an ellipsoid, justify why a minimum exists in Ω, manipulate gradients of quadratic forms in matrix notation, and keep track of operator inverses – all absent from the original two-dimensional, isotropic setting.\n\nHence the enhanced variant is markedly more intricate, demands broader mathematical tools, and carries a non-trivial optimality component, making it substantially harder than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}