{ "Tag": [ "inequalities", "probability", "induction", "function", "probability and stats" ], "Problem": "Hello. Our lecturer stated the Bonferroni Inequalites as follows:\r\n\r\nLet $ A_1, A_2, A_3 \\ldots$ be events, and $ \\cal P$ be a probability distribution. Let $ \\cal A$ be an odd/even down-set consisting of subsets of $ \\{1, 2, 3, \\ldots, n\\}$. Let $ A_X = \\bigcap_{x \\in X} A_x$. Then:\r\n\r\n${ {\\cal P} \\left(A_1 \\cup \\ldots \\cup A_n \\right) \\leq / \\geq \\sum_{X \\in {\\cal{A}}, X \\neq \\phi} (-1)^{|X| + 1} {\\cal{P}} (A_X)}$\r\n\r\nIs this correct, as when he comes to prove it, he only considers down-sets with only one maximilised set? (i.e. the power set of a subset of $ \\{1, 2, 3, \\ldots, n\\}$). If it is correct, can you provide a proof for me...it probably can be done by induction, but I am too lazy to construct a proof myself.", "Solution_1": "P.S. our lecturer is the esteemed field medalist Timothy Gowers.", "Solution_2": "I was under the impression he retracted his claim of the theorem as you have stated it; Imre Leader appeared to think there was something wrong with it...", "Solution_3": "Oh ok...thanks...I never turn up for lectures...", "Solution_4": "i gave a very handwavy proof of what are usually called the bonferroni inequalities [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169137]here[/url].\r\n\r\n(the idea is to prove the corresponding pointwise inequality for the indicator functions of the $ A_i$s and use that $ 1_{A_i\\cap A_j}\\equal{}1_{A_i}1_{A_j}$ + a bit of elementary combinatorics)" } { "Tag": [ "geometry unsolved", "geometry" ], "Problem": "I want a pure geometric solution for the following question\r\n\r\n1)If $ BC$ is the greatest side of $ \\triangle ABC$ and $ D, E$ are points on $ BC, CA$ respectively prove that $ BC \\ge DE$\r\n\r\n2)If $ BC$ is the greatest side of $ \\triangle ABC$ and $ E, F$ are points on $ AB, CA$ respectively prove that $ BC \\ge EF$\r\n\r\n3)Prove that no straight line can be drawn within a triangle which is greater than the greatest side.\r\n\r\nI want separate solutions but its Ok if u just give one solution for the first two \r\nI'll work it out", "Solution_1": "That's a simple problem.\r\nIn problem a-) (using the same strategy again in the other items) see when $ \\angle DEA$ is obtuse or not, spliting in some cases. Like this you will have an easy but non-beautiful solution.", "Solution_2": "I dont understand\r\nIts trivial that when $ \\angle EDC \\ge 90$ then $ ED < BC$ but what if it is acute?? :?: \r\nAlso I understand that when the $ \\angle EDC$ tends to $ 0$ then it become equal....\r\nPlease help me out....... :maybe: :(", "Solution_3": "$ \\angle DEA$ is obtuse $ \\Rightarrow DA \\geq DE$, and $ DA \\le AB$ or $ AC$, and both are less than or equal to $ BC$.\r\nNow $ \\angle DEA$ is acute $ \\Rightarrow DE \\le DC \\le BC$, and it's done.", "Solution_4": "Thanx erudito for all ur help\r\n\r\ncan someone help me on these questions\r\nhttp://www.mathlinks.ro/viewtopic.php?t=248523\r\nhttp://www.mathlinks.ro/viewtopic.php?t=247475\r\nhttp://www.mathlinks.ro/viewtopic.php?t=247471\r\n\r\nplease i need all the help u can give me..........." } { "Tag": [ "calculus", "derivative", "function", "inequalities", "complex analysis", "complex analysis unsolved" ], "Problem": "Let $ H\\equal{}\\{ z: \\Im (z) \\ge 0 \\}$ denote the upper half plane. Suppose $ F: H \\to H$ is analytic and $ a \\in H$. Prove that $ |F'(a)| \\le \\frac{ \\Im F(a)}{ \\Im a}$.", "Solution_1": "I only can take a quick guess because I don't have time to check it and even if it's wrong you can solve it as an exercise. $ H$ and the unit disk are the same, I mean there's a suitable Mobius fractional function which transforms one of them into other and in unit disk we have a very useful inequality, known as the Pick's lemma which states, $ \\frac {|f'(z)|}{1 \\minus{} |f(z)|^2} \\leq \\frac {1}{1 \\minus{} |z|^2}$ and the equality holds when $ f$ is a Mobius function and I think, it should derive from this lemma. :wink:", "Solution_2": "Take a conformal map $ D\\to H$ that sends $ 0$ to $ a$. \r\nFollow it with $ F$.\r\nThen map back onto the disk, sending $ F(a)$ to $ 0$. Now the ordinary Schwarz lemma (rather than Schwarz-Pick) can be applied.\r\n\r\nOf course, Schwarz-Pick is just Schwarz augmented by a conformal mapping." } { "Tag": [ "inequalities" ], "Problem": "$ a^{3}\\plus{}b^{3}\\plus{}c^{3} \\geq a^{2}b\\plus{}b^{2}c\\plus{}c^{2}a$\r\nfor nonnegative a, b, c.", "Solution_1": "Consider the sequences:\r\n\r\n$ a,b,c$, and $ a^2,b^2,c^2$\r\n$ b,c,a$ and $ a^2$, $ b^2$, $ c^2$\r\n\r\nIf either $ a\\ge b\\ge c$ or $ c\\ge b\\ge a$, the top sequences are like sorted, and the bottom array isn't, so we are done by the rearrangement inequality.", "Solution_2": "Well for positive $ a,b,c$, $ a,b,c$ and $ a^2,b^2,c^2$ are always similarly sorted, and are bigger than any other scalar product of them...\r\n\r\nIn general, I think\r\n\r\n$ a^k\\plus{}b^k\\plus{}c^k \\ge \\sum_{cyc}{a^{k_1}b^{k_2}c^{k_3}}$ for $ k\\equal{}k_1\\plus{}k_2\\plus{}k_3$\r\n\r\nholds for positive $ a,b,c,k$.", "Solution_3": "You can do rearrangement on three sequences too so that proves your generalization (as well as an n variable one).", "Solution_4": "[quote=\"not_trig\"]Well for positive $ a,b,c$, $ a,b,c$ and $ a^2,b^2,c^2$ are always similarly sorted, and are bigger than any other scalar product of them...\n\nIn general, I think\n\n$ a^k \\plus{} b^k \\plus{} c^k \\ge \\sum_{cyc}{a^{k_1}b^{k_2}c^{k_3}}$ for $ k \\equal{} k_1 \\plus{} k_2 \\plus{} k_3$\n\nholds for positive $ a,b,c,k$.[/quote]\r\n\r\nIf $ k_1,k_2,k_3$ are nonnegative, of course. With AM-GM, we get $ k_1a^k\\plus{}k_2b^k\\plus{}k_3b^k\\geq ka^{k_1}b^{k_2}c^{k_3}$ etc.\r\n\r\nIf one is negative, of course not (consider $ a\\equal{}b\\equal{}1,c\\equal{}\\epsilon$ for $ \\epsilon$ really close to 0)." } { "Tag": [ "geometry", "trapezoid", "geometry open" ], "Problem": "Let $ ABCD$ be a quadrilateral and let $ P$ be the intersection of $ AB$ and $ CD$. For any $ R\\in [AD]$ let $ f(R) \\in [CD]$ be the intersection between $ PR$ and $ CD$. What is the locus of the midpoints of segments $ Rf(R)$?", "Solution_1": "[quote=\"mychrom\"]Let $ ABCD$ be a quadrilateral and let $ P$ be the intersection of $ AB$ and $ CD$. For any $ R\\in [AD]$ let $ f(R) \\in [CD]$ be the intersection between $ PR$ and $ CD$. What is the locus of the midpoints of segments $ Rf(R)$?[/quote]\r\n\r\nI think that $ f(R)\\in [BC].$ It's an interesting problem. If you heard about that problem from a book the locus can be a circle (if $ ABCD$ is a trapezoid then the radius of the circle is $ r\\equal{}\\infty$). You may try to prove that points $ M,K_{1},K,N$, from my figure are concyclic using complex numbers. The locus we're looking for is marked with [color=red]red[/color].\r\n\r\n\r\n[url=http://imageshack.us][img]http://img178.imageshack.us/img178/3999/maximbogdan2xh9.png[/img][/url]" } { "Tag": [ "trigonometry", "integration", "algebra", "linear equation", "calculus", "calculus computations" ], "Problem": "(1) Solve $ t^2\\frac {dx}{dt} \\minus{} 2tx \\equal{} 3$.\r\n\r\n(2) Show that $ \\frac {dy}{dt} \\plus{} P(t)y \\equal{} Q(t)y^n$ will become a linear equation by proper transformation.\r\n\r\n(3) In (2), find the solution for $ P(t) \\equal{} \\frac {2}{t},\\ Q(t) \\equal{} t^2\\sin t,\\ n \\equal{} 2$.", "Solution_1": "For the first one:\r\n$ \\displaystyle t^2\\frac {dx}{dt} \\minus{} 2tx \\equal{} 3$\r\n$ \\displaystyle \\frac {dx}{dt} \\equal{} \\frac {3}{t^2} \\plus{} 2\\frac {x}{t}$\r\nLet $ v \\equal{} x/t$ then $ \\displaystyle \\frac {dx}{dt} \\equal{} v \\plus{} t\\frac {dv}{dt}$\r\nThus, $ \\displaystyle v \\plus{} t\\frac {dv}{dt} \\equal{} \\frac {3}{t^2} \\plus{} 2v\\Rightarrow t\\frac {dv}{dt} \\equal{} \\frac {3}{t^2} \\plus{} v\\Rightarrow \\frac {dx}{dt} \\minus{} \\frac {v}{t} \\equal{} \\frac {3}{t^3}...(1)$\r\nThe integrating factor is $ \\displaystyle e^{\\int \\minus{} \\frac {1}{t}dt} \\equal{} \\frac {1}{t}$\r\nTake the differential of $ (1)$, and multiply both sides of the resulting equation by $ 1/t$.\r\nThen integrate to get $ \\displaystyle \\int d(v/t) \\equal{} \\int\\frac {3}{t^4}dt\\Rightarrow v \\equal{} \\minus{}t^{\\minus{}2}\\plus{}Ct$\r\nTherefore, $ \\displaystyle x \\equal{} \\minus{} \\frac {1}{t} \\plus{} Ct^2$.", "Solution_2": "2) $ \\dfrac{\\mathrm{d}y}{\\mathrm{d}t} \\plus{} P(t)y\\equal{}Q(t)y^n$\r\nDivide throughout by $ y^n$ to get \r\n$ \\dfrac{1}{y^n}\\cdot\\dfrac{\\mathrm{d}y}{\\mathrm{d}t} \\plus{} \\dfrac{1}{y^{n\\minus{}1}}P(t)\\equal{}Q(t)$\r\nSet $ \\dfrac{1}{y^{n\\minus{}1}}\\equal{}u\\ \\Rightarrow \\ \\dfrac{1}{y^n}\\,\\dfrac{\\mathrm{d}y}{\\mathrm{d}t}\\equal{}\\dfrac{1}{1\\minus{}n}\\cdot\\dfrac{\\mathrm{d}u}{\\mathrm{d}t}$\r\nThus the given equation\r\n$ \\dfrac{1}{1\\minus{}n}\\cdot \\dfrac{\\mathrm{d}u}{\\mathrm{d}t}\\plus{}P(t)u \\equal{} Q(t)$\r\nwhich is the same as the linear equation \r\n$ \\dfrac{\\mathrm{d}u}{\\mathrm{d}t}\\plus{}(1\\minus{}n)P(t)u \\equal{} (1\\minus{}n)Q(t)$\r\nP.S: This is the Bernoulli's eqn.", "Solution_3": "Those are correct. How about (3)?", "Solution_4": "3) Setting $ P(t) \\equal{} \\dfrac{2}{t}$, $ Q(t)\\equal{}t^2\\sin t$ and $ n\\equal{}2$, we get, after setting $ u\\equal{}\\dfrac{1}{y}$, \r\n$ \\dfrac{\\mathrm{d}u}{\\mathrm{d}t} \\minus{}\\dfrac{2u}{t}\\equal{}\\minus{}t^2\\sin t$\r\n$ \\dfrac{1}{t^2}\\cdot\\dfrac{\\mathrm{d}u}{\\mathrm{d}t} \\minus{}\\dfrac{2u}{t^3}\\equal{}\\minus{}\\sin t$ (multiplying throughout by $ \\dfrac{1}{t^2}$)\r\n$ \\Rightarrow\\ \\dfrac{\\mathrm{d}}{\\mathrm{d}t}\\left(\\dfrac{u}{t^2}\\right) \\equal{} \\minus{}\\sin t$\r\n$ \\Rightarrow \\dfrac{u}{t^2} \\equal{} \\cos t \\plus{}C$\r\n$ \\Rightarrow\\ y \\equal{} \\dfrac{1}{t^2\\cos t \\plus{} Ct^2}$", "Solution_5": "That's right. :)" } { "Tag": [], "Problem": "Any trivia buffs out there? IM me at P3nguinP0wered on AIM and we can play!!", "Solution_1": "go to:\r\n\r\nhttp://www.funtrivia.com/private/main.cfm?tid=960\r\n\r\nAoPS has it s own trivia thing" } { "Tag": [ "integration", "calculus", "function", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "$\\int_{\\pi^{e}}^{e^{\\pi}}\\frac{dt}{\\sqrt{\\zeta^{3}(x^{2}(t))-1}}\\int_{\\pi^{e}}^{e^{\\pi}}\\sqrt{\\zeta^{3}(x^{2}(t))-1}dt\\to min$ if\r\n$2\\leq x(t) \\leq 4$\r\nfind also for wich $x(t)$ hold equality.", "Solution_1": "By Caushy-Schwarz, the product of the integrals is never less than $(b-a)^{2}$ where $a=\\pi^{e}$ and $b=e^\\pi$ are the limits of integration. Any $x$ that makes the integrands constant functions gives an equality, so the answer to the second part is any constant function $x(t)=c\\in[2,4]$. Again, I don't quite understand what's the point in piling up $\\zeta$'s, squares and square roots. There is no real difficulty in the question, just a lot of irrelevant notation. I could also ask: what's the minimum of\r\n\\[\\left(x\\frac{e^{e^{x-239}}+x^{2}\\log(1+x^{32})}{\\sin(cos(e^{x}))+2}\\right)^{2002}\\]\r\nbut what's the point? The problem basically reduces to the fact that every square is non-negative and $0$ is attained. Let me make myself even more clear: if you continue to post the problems, which, stripped of all fancy notation and other irrelevant things, are trivial, nobody will read your posts. A good problem has no added artificial complexities in it, the intrinsic difficulty of the question should be enough to make it interesting. I loved your question about $\\sum_{k}(-1)^{[kx]}$ but what you posted lately is not really what I would call attractive and, judging from other people's lack of response, almost no one else seems to get interested. :(", "Solution_2": "Sorry, I don't posted right my problem. You must find $max$ .\r\nI mean: \r\n\r\n$\\int_{\\pi^{e}}^{e^{\\pi}}\\frac{dt}{\\sqrt{\\zeta^{3}(x^{2}(t))-1}}\\int_{\\pi^{e}}^{e^{\\pi}}\\sqrt{\\zeta^{3}(x^{2}(t))-1}dt\\to max$ if\r\n$2\\leq x(t) \\leq 4$\r\nfind also for wich $x(t)$ hold equality.\r\nyou only need to use http://www.mathlinks.ro/Forum/viewtopic.php?t=135955\r\nI understand that my problems is not so interesting because it's based on combinations Holder's inequality and etc... sorry\r\n\r\n\r\n. \r\n\r\nThanks for your comment fedja." } { "Tag": [], "Problem": "if I have f(x)=x^4 +x +3\r\nThey asked me find the f(-x), what is the result of it, and why? if you please giving me the formulas to find f(-x).", "Solution_1": "Here you are. Simply to find f(-x) you have to change x into -x and simplify the equation.\r\nIf f(x)=x^4 +x +3 than f(-x)=(-x)^4+(-x)+3=x^4-x+3. Good luck!", "Solution_2": "i confused...\r\n\r\nis the answer X^4-X+3, or not...\r\n\r\nbecause the problem isnt hard,and i think,the problem that Camytran said,is another thing...and need to be\r\n\r\n discussed... :)", "Solution_3": "[quote=\"ashegh\"]i confused...\n\nis the answer X^4-X+3, or not...\n\nbecause the problem isnt hard,and i think,the problem that Camytran said,is another thing...and need to be\n\n discussed... :)[/quote]\r\n\r\nYes, you're rite, the problem is not easy to solve like that, because the answer teacher gives me is differerent, and i dunno y it's like that. Can u give me the other solution?", "Solution_4": "ok ok ,i will do my best...\r\n\r\nbut could u plz tell me what did u teacher say about it exactly?\r\n\r\nand could u plz say again,what the problem want me to do?\r\n\r\nbecause i want to be sure about it ;) .", "Solution_5": "[quote=\"ashegh\"]\n\nis the answer X^4-X+3, or not...\n\n[/quote]\r\n\r\nYes, it is! :)\r\n\r\npupOK's method is the good one...", "Solution_6": "[quote=\"ashegh\"]ok ok ,i will do my best...\n\nbut could u plz tell me what did u teacher say about it exactly?\n\nand could u plz say again,what the problem want me to do?\n\nbecause i want to be sure about it ;) .[/quote]\r\n\r\nHey, if u dun mind, can u give me your y!m nick? so i can tell you clearer and faster.", "Solution_7": "[quote=\"camytran\"][quote=\"ashegh\"]ok ok ,i will do my best...\n\nbut could u plz tell me what did u teacher say about it exactly?\n\nand could u plz say again,what the problem want me to do?\n\nbecause i want to be sure about it ;) .[/quote]\n\nHey, if u dun mind, can u give me your y!m nick? so i can tell you clearer and faster.[/quote]\r\nI know that you are new, but next time try not to use the phrases: \"u\" instead of \"you\" and \"dun\" instead of \"don't\" becuase it just makes reading it a whole lot easier. :)", "Solution_8": "yea sometimes people mis read things like u might be a variable and you would use it to say you so people would get confused", "Solution_9": "[quote=\"camytran\"][quote=\"ashegh\"]ok ok ,i will do my best...\n\nbut could u plz tell me what did u teacher say about it exactly?\n\nand could u plz say again,what the problem want me to do?\n\nbecause i want to be sure about it ;) .[/quote]\n\nHey, if u dun mind, can u give me your y!m nick? so i can tell you clearer and faster.[/quote]\r\n\r\nyou can look at his profile or the bottom of his post. Look for \"YIM\".", "Solution_10": "I don't think there's any other solution! The mean I proposed is the easiest. And what answer were you given?", "Solution_11": "the problem is too easy,and dont need to be discussed more..." } { "Tag": [ "geometry", "geometric transformation", "reflection", "algebra proposed", "algebra" ], "Problem": "Solve in real numbers: $6^{x}+1=8^{x}-27^{x-1}$", "Solution_1": "This is from Mathematical Reflections, see [url]http://reflections.awesomemath.org/2006_2/2006_2_solutions.pdf[/url]", "Solution_2": "Hmm... has it appeared somewhere before?" } { "Tag": [ "calculus", "integration", "complex analysis" ], "Problem": "Hi, \r\n\r\nHow would you show that if $ \\mathcal{F}\\subset H(G)$ is normal then $ \\mathcal{F'}\\equal{}\\{f': f\\in\\mathcal{F}\\}$ is also normal?\r\n\r\nI'm assuming that I need to use Montel's theorem, but I don't know how to. \r\n\r\nThanks for the help.", "Solution_1": "You have locally uniform estimates for $ f$, and you want locally uniform estimates for $ f'$. The Cauchy integral formula is the key here.\r\n\r\nWe can be more direct than Montel's theorem here: if $ f_n$ converges uniformly on compact sets to a limit $ f$, $ f_n'$ converges uniformly on compact sets to $ f'$. That's enough convergent sequences right there." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Let $ a,b,c\\in Z_\\plus{}$ such that $ a^3\\plus{}b^3\\plus{}c^3\\equal{}d^3$. Does there exist $ n>3$ and $ n\\in Z_\\plus{}$ such that\r\n$ a^n\\plus{}b^n\\plus{}c^n\\equal{}d^n$", "Solution_1": "$ d^{3\\plus{}n} \\equal{} a^3d^n \\plus{} b^3d^n \\plus{} c^3d^n > a^{3\\plus{}n} \\plus{} b^{3\\plus{}n} \\plus{} c{3\\plus{}n}$ because $ d>a$, $ d>b$, $ d>c$." } { "Tag": [ "inequalities" ], "Problem": "This problem is from Olimpiads in 2004 from Romania!\r\n Find all real numbers x,y,z sug as :\r\n x+y+z=3 and x^8+y^8+z^8=3", "Solution_1": "C'mon guys, this is so easy!\r\nWhat about Cauchy (or, even better, the power mean inequality)? :D", "Solution_2": "By Power Mean, ((x^8+y^8+z^8)/3)^(1/8) >= (x+y+z)/3, but here we have equality. Therefore, x=y=z=1.", "Solution_3": ":D :D :D Hehe Arne getting impatient on the beginners. Very unlike him..." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "All integer numbers [tex]c[/tex],give equation:[tex]3y^2+4cy^3+2xy+48=0[/tex].Find all c in Z such that:the amount of solution [tex](x,y)[/tex] satisfying A,B conditions is maximum:\r\nA.\\x\\ is perfect square\r\nB.there is no integer p such that:[tex]p^2[/tex] divides [tex]y[/tex]", "Solution_1": "No solution for it?? :(", "Solution_2": "Oh.No solution....after 3 months...." } { "Tag": [ "AMC", "AMC 8", "MATHCOUNTS", "geometry", "induction" ], "Problem": "I am homeschooling a middle school student who is trying to increase his speed at solving math problems. He knows the concepts and can answer all the problems he faces on tests such as the AMC8, SATs, and MathCounts, but he's too slow and methodical to finish within the allotted time. This is keeping him from scoring beyond a 700 on the SAT or making it to countdown in MathCounts because he doesn't finish all the problems to get the top scores.\r\n\r\nDoes anyone have suggestions to help him? He's almost finished with the Intro to Geometry book and has completed all the other Intro books. Again, he doesn't need help with concepts and basic math, just techniques to improve his speed.\r\n\r\nThanks!", "Solution_1": "Don't second guess what you're doing. If you carry out a computation, assume it's right unless you have reason not to. If you spend too much time re-checking what you're doing before you've had a chance to complete other problems, you're going to waste time. Don't bother checking yourself until you've answered as many problems as you can. Maybe some people have other strategies but this helps me go faster.", "Solution_2": "I think that if you just do really really a lot of practice, your speed will increase. Because when you do lots of problems, you gain experience, and your intuition gets better, as well as you get more used to doing routine calculations (or parts of proofs that consistently occur, e.g. induction, standard combinatorial sums). As a result, when you you see a problem, you will be thinking faster and it is more likely you will be thinking along the right lines, as a result you will come up with the solution faster.\r\nAs for techniques, again, if you do a lot of problems, your will learn more techniques as well as figure out which techniques are useful to remember and you will master them better. For checking, I find it difficult to check the first few problems after going through the whole test, as I sometimes forget what I did and need to resolve it. Maybe an efficient idea is checking your answers/solutions after doing a certain number of problems, so that the solution to them is still fresh in your head (e.g. after doing 5 problems, check them, and then move on, on the AMC. Depending on your preferences, this number could be less or greater. You can figure out what works best for you again by trying different styles on past contests).\r\nHowever, this was just my experience, I am not sure how efficient this is. Because the amount of practice you need for certain things to get really good at them, is sometimes quite large, as a result when you compare your speed at the moment to what it was a week ago, there will probably not be much change. But if it is compared to a year ago, and if you have consistently practiced, then yeah, it will probably show a pretty significant increase." } { "Tag": [ "inequalities", "inequalities proposed" ], "Problem": "Let $ a,b,c$ be nonnegative real numbers, prove that\r\n\\[ \\sqrt {\\frac {8a^2 \\plus{} bc}{b^2 \\plus{} c^2}} \\plus{} \\sqrt {\\frac {8b^2 \\plus{} ca}{c^2 \\plus{} a^2}} \\plus{} \\sqrt {\\frac {8c^2 \\plus{} ab}{a^2 \\plus{} b^2}} \\ge \\frac {9}{\\sqrt {2}}\r\n\\]\r\nEquality holds for $ a \\equal{} b \\equal{} c$ or $ a \\equal{} b,c \\equal{} 0$ or any cyclic permutations. :)\r\n\r\nPS: This is a problem I posted in the topic \"Some new results\". I like it very much. I hope we will discuss more about it. Thank you very much! :)", "Solution_1": "[quote=\"can_hang2007\"]Let $ a,b,c$ be nonnegative real numbers, prove that\n\\[ \\sqrt {\\frac {8a^2 \\plus{} bc}{b^2 \\plus{} c^2}} \\plus{} \\sqrt {\\frac {8b^2 \\plus{} ca}{c^2 \\plus{} a^2}} \\plus{} \\sqrt {\\frac {8c^2 \\plus{} ab}{a^2 \\plus{} b^2}} \\ge \\frac {9}{\\sqrt {2}}\n\\]\nEquality holds for $ a \\equal{} b \\equal{} c$ or $ a \\equal{} b,c \\equal{} 0$ or any cyclic permutations. :)\n\nPS: This is a problem I posted in the topic \"Some new results\". I like it very much. I hope we will discuss more about it. Thank you very much! :)[/quote]\r\n[hide=\"hint\"]\\[ bc \\ge \\frac{2b^2c^2}{b^2\\plus{}c^2}\\] :)[/hide]", "Solution_2": "Did you use Holder in your solution and p,q,r method?\r\n\r\nThank you very much.\r\n\r\n\r\nP.S. Really a beautiful inequality :wink:" } { "Tag": [ "function", "complex analysis", "complex analysis unsolved" ], "Problem": "Hi,\r\nsuppose $ L\\subset\\mathbb C$ is a lattice.\r\n$ f, g$ are two elliptic functions with $ L$ as periodic lattice.\r\n\r\nHow to show that\r\ni) if $ f$ and $ g$ have the same poles and the same principal part, \r\nthen $ f\\equal{}g\\plus{}c$ for a constant $ c$.\r\n\r\nand\r\n\r\nii) if $ f$ and $ g$ have got the same poles and zeros (with the same multiplicities), \r\nthen $ f\\equal{}cg$ for a constant $ c$.", "Solution_1": "In case i), what properties does $ f\\minus{}g$ satisfy? What do you know about elliptic functions with these properties?\r\n\r\nAlso do the same for $ \\frac{f}{g}$.", "Solution_2": "[quote=\"calc rulz\"]In case i), what properties does $ f \\minus{} g$ satisfy? What do you know about elliptic functions with these properties?\n\nAlso do the same for $ \\frac {f}{g}$.[/quote]\r\n\r\nHi,\r\ni get it. :lol: thanks very much." } { "Tag": [ "calculus", "integration", "real analysis", "function", "real analysis theorems" ], "Problem": "I'm hoping someone can help me address this query: How many different ways are there of introducing Measure Theory? I am aware of the following 3 methods, and wonder if these are it, or there are more.\r\n\r\n1. Paul Halmos' approach using measures on rings of sets\r\n2. Walter Rudin's approach using linear functionals\r\n3. Daniell's approach\r\n\r\nI hope to hear soon. Thanks!", "Solution_1": "Most of books in China use (1) and (maybe therefore) I feel that it is easier to understand than (2).\r\n\r\nWhat is (3)?", "Solution_2": "Does anyone know some good books with theory and also with exercices for Measure Theory? ;)", "Solution_3": "I like Rudin's Real and Complex Analysis.", "Solution_4": "In general, any book with \"Real Analysis\" in the title is likely to be useful. The details will very, since there are a lot of these textbooks, but all should have measure theory.", "Solution_5": "I'm partial to Wheeden and Zygmund, [i]Measure and Integral[/i]. The approach there is to work out from the beginning the details of Lebesgue measure in $\\mathbb{R}^n.$ This differs on the one hand from approaches that start with just $\\mathbb{R}$ and on the other hand from approaches that start with abstract measure spaces.\r\n\r\nThe integral of a function $f: E\\mapsto [0,\\infty]$ where $E\\subset\\mathbb{R}^n$ is then defined as the $(n+1)$-dimensional measure of the set $\\{(x,y): x\\in E,0\\le y \\le f(x)\\}.$ One advantage of this is that it produces Fubini's theorem earlier and more naturally." } { "Tag": [ "number theory unsolved", "number theory" ], "Problem": "Given the equation $ x^2\\plus{}y^2\\plus{}z^2\\plus{}t^2\\minus{}nxyzt\\minus{}n\\equal{}0$ where $ x,y,z,t$ are positive integers.Prove that this equation is solvable when $ n\\equal{}2008$ and isn't solvable when $ n\\equal{}2007$", "Solution_1": "For $ n \\equal{} 2008$, try $ x \\equal{} y \\equal{} 6$ and $ z \\equal{} 44$." } { "Tag": [ "inequalities unsolved", "inequalities" ], "Problem": "Let $ a,b,c\\geq0$ s.t. $ a\\plus{}b\\plus{}c\\equal{}3$.Prove that:\r\n$ \\dfrac{1}{2ab^2\\plus{}1}\\plus{}\\dfrac{1}{2bc^2\\plus{}1}\\plus{}\\dfrac{1}{2ca^2\\plus{}1}\\geq1$.", "Solution_1": "[quote=\"Inequalities Master\"]Let $ a,b,c\\geq0$ s.t. $ a \\plus{} b \\plus{} c \\equal{} 3$.Prove that:\n$ \\dfrac{1}{2ab^2 \\plus{} 1} \\plus{} \\dfrac{1}{2bc^2 \\plus{} 1} \\plus{} \\dfrac{1}{2ca^2 \\plus{} 1}\\geq1$.[/quote]\r\n$ <\\equal{}>(ab^2\\plus{}bc^2\\plus{}ca^2)\\plus{}1 \\ge 4 (abc)^3$ ; $ t \\equal{} abc \\le 1$;\r\n$ ab^2\\plus{}bc^2\\plus{}ca^2 \\ge 3t$ ; $ 3t\\plus{}1 \\ge 4t^3$ $ (0 \\le t \\le 1)$;", "Solution_2": "[quote=\"Inequalities Master\"]Let $ a,b,c\\geq0$ s.t. $ a \\plus{} b \\plus{} c \\equal{} 3$.Prove that:\n$ \\dfrac{1}{2ab^2 \\plus{} 1} \\plus{} \\dfrac{1}{2bc^2 \\plus{} 1} \\plus{} \\dfrac{1}{2ca^2 \\plus{} 1}\\geq1$.[/quote]\r\n\r\n\r\n$ \\frac{1}{2ab^{2}\\plus{}1}\\equal{}1\\minus{}\\frac{2ab^{2}}{2ab^{2}\\plus{}1}\\equal{}1\\minus{}\\frac{2ab^{2}}{ab^{2}\\plus{}ab^{2}\\plus{}1}\\geq 1\\minus{}\\frac{2ab^{2}}{3\\sqrt[3]{a^{2}b^{4}}}\\equal{}1\\minus{}\\frac{2\\sqrt[3]{a^{2}b}}{3}\\geq 1\\minus{}\\frac{2(a\\plus{}a\\plus{}b)}{9}\\equal{}1\\minus{}\\frac{2(2a\\plus{}b)}{9}$\r\n\r\n$ \\dfrac{1}{2ab^2 \\plus{} 1} \\plus{} \\dfrac{1}{2bc^2 \\plus{} 1} \\plus{} \\dfrac{1}{2ca^2 \\plus{} 1}\\geq 1\\minus{}\\frac{2(2a\\plus{}b)}{9}\\plus{}1\\minus{}\\frac{2(2b\\plus{}c)}{9}\\plus{}1\\minus{}\\frac{2(2c\\plus{}a)}{9}\\equal{}3\\minus{}\\frac{2}{9}(3a\\plus{}3b\\plus{}3c)\\equal{}3\\minus{}2\\equal{}1$\r\n\r\n$ a\\equal{}b\\equal{}c\\equal{}1$" } { "Tag": [ "induction" ], "Problem": "There are 25 stones in a heap. The heap is divided into two parts, then one of the parts is divided in two again,etc,until we have 25 separate stones.After each division of one of the heaps into two smaller heaps we write the product of the numbers of stones in these two heaps .Prove that at the end the sum of all the numbers is 300.", "Solution_1": "[hide=\"Easy\"]Use Second Principle of Finite Induction, knowing the result for number of stones n=1,2,3 and use $ f(n)\\equal{}\\frac {n(n\\minus{}1)}{2}$[/hide]" } { "Tag": [ "MATHCOUNTS", "articles", "probability", "percent", "geometry", "3D geometry", "dodecahedron" ], "Problem": "16 hours after the scheduled end of the Countdown Round, and still no results posted here or on the MATHCOUNTS Web Site. I find an online news article that reports:\r\n\r\nIllinois Eighth - Grader and Illinois Team Awarded Mathematics Champions at 2004 MATHCOUNTS National Competition -- Friday May 7, 10:06 pm ET[quote]Gregory Gauthier was victorious in the intense, one-on-one oral Countdown Round where the top 12 students competed for the National Championship title. Jeffrey Chen of College Station, Tex., was awarded the second-place individual title with Haitao Mao of Falls Church, Va. and Sam Elder of Ft. Collins, Colo., advancing to the Semi-finals.[/quote]\r\n\r\nI hope those \"in the know\" will provide further details. Thanks!", "Solution_1": "I have heard just a bit more through the grapevine - consider none of these results 'official'.\r\n\r\nI heard that Greg (IL) won both the written round and the countdown round.\r\n\r\nIllinois won the Team competition with all 4 students probably ranked 15th or better. The Individual (combined) score for 15th was around 36. \r\n\r\nAny other news out there?", "Solution_2": "California was second-place team. New York was sixth place team and Danny Zhu from NY was seventh place individual.", "Solution_3": "I too heard that Gregory Gauthier won.", "Solution_4": "I got this site from Alison:\r\n\r\nhttp://biz.yahoo.com/prnews/040507/lnf007_1.html", "Solution_5": "Didn't he do well in last year's competition? I think I've heard his name somewhere before...", "Solution_6": "ya, he was in the top 10 last year too (figures that he'd win this year). didn't know that he was a 7th grader last year though.", "Solution_7": "yeah, I'd thought Shaunak was the only sevie last yr...", "Solution_8": "maybe he stayed back a grade to win MC :P", "Solution_9": "lol...i've thought about doing that myself...lol...", "Solution_10": "yeah, greg won everything. everything. written round, final standings, masters, team, everything.", "Solution_11": "rcv - Sam did Colorado proud. He finished 12th on the written and mowed through the countdown round until running into Greg in the semis.\r\n\r\nGreg put on one of the most impressive shows I've ever seen at Nationals.\r\n\r\nSeveral AoPSers made the top 12, and I was very happy to meet many of you there (though I wish I'd had more time to talk to you all!)", "Solution_12": "According to the [url=http://mathcounts.org/Competitions/2003StateWinners.html]2003 state winners list[/url], Greg Gauthier was a 7th grader last year. He did not stay back an extra year. He was just a three-year competitor, I think.\r\n\r\nI seem to remember he placed in 6th grade also. Or no. He was just on the 2nd place Illinois team with Albert Ni.", "Solution_13": "I vaguely remember that the ESPN broadcast last year mistakenly said that Greg was an 8th grader that year. But I could be the one that is mistaken.", "Solution_14": "5th place written here. Got booted first round of countdown. I also would've been 2nd written if only I had realized that the question asked for a percent, not a common fraction. Oh well, MC is done for me!", "Solution_15": "Wasn't #2 about the dodecahedron net and the shortest distance between a and b? The answer to that was 3.", "Solution_16": "yea i missed it i got 4....", "Solution_17": "[quote=\"captcha000\"]yea i missed it i got 4....[/quote]\r\nlol... so did i.\r\ncan't believe that was the only target i missed", "Solution_18": "i talked to so many people who got 4 for that one including two of my teammates.", "Solution_19": "How do you do that one. -1 :le: x :le: 3\r\nand -2 :le: y :le: 4? \r\n\r\nI keep getting 14/35. \r\n\r\nFor x, there is -1,0,1,2,3 and for y there's -2,-1,0,1,2,3,4. So that is 5*7=35 pairs of xy. How can you get 12 as the denominator?", "Solution_20": "It's for reals x and y.", "Solution_21": "[quote=\"white_horse_king88\"]How do you do that one. -1 :le: x :le: 3\nand -2 :le: y :le: 4? \n\nI keep getting 14/35. \n\nFor x, there is -1,0,1,2,3 and for y there's -2,-1,0,1,2,3,4. So that is 5*7=35 pairs of xy. How can you get 12 as the denominator?[/quote]\r\nthat's exactly what i did. i thought it meant integers so i got screwed there", "Solution_22": "I got 98th, but Nebraska got 12th because we had the 19th place kid (alc) and we got a 10 on team.", "Solution_23": "10 on team is vital for a good team ranking...", "Solution_24": "Results are posted. [url]http://mathcounts.org/Competitions/NatlCompRankings04.pdf[/url]", "Solution_25": "[quote=\"Syntax Error\"]Well, the thinking was more of a \"there isn't much of a difference between 1st and 10th place written\" because usually they aren't separated by too many questions.\n\nPersonally, I favor the new countdown system.[/quote]\r\n\r\nI haven't been on this forum for a whole week. WHat's the new countdown system?", "Solution_26": "http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12041", "Solution_27": "Well... did well on sprint, but totally bombed the target... so i got 24 + 6 = 30... (51st) \r\n\r\nOK got 18th... 10 higher than last year...", "Solution_28": "[quote=\"Chinaboy\"]Well... did well on sprint, but totally bombed the target... so i got 24 + 6 = 30... (51st) \n\nOK got 18th... 10 higher than last year...[/quote]\r\n\r\nHuh? You got 18th, 28th, and 51st somehow... and it seems that two are in the same year...\r\n\r\nEdit: Ahhh... OK = Oklahoma... makes sense now ;-).", "Solution_29": "???\r\n\r\nOK = Oklahoma\r\n\r\nhopes that clears everything up" } { "Tag": [ "geometry", "3D geometry" ], "Problem": "Find the surface area of a cube with a length of 10, and a height of 10. (is that enough information to solve it?)", "Solution_1": "[hide=\"my answer\"]\n10*10*6=600[/hide]", "Solution_2": "10 x 10 x 6 = 600 units squared", "Solution_3": "600", "Solution_4": "actually the information is more than enough to solve it!", "Solution_5": "i forgot how to do surface area, and i didn't know what you needed", "Solution_6": "600", "Solution_7": "[hide]yes, it is more information than you need. Since the question states that it is a cube, then you wouldn't need both the width and the height, just one is enough. 10 x 10 x 6 = 600 sq units[/hide]", "Solution_8": "[quote=\"BOOYA\"][hide]yes, it is more information than you need. Since the question states that it is a cube, then you wouldn't need both the width and the height, just one is enough. 10 x 10 x 6 = 600 sq units[/hide][/quote]\r\n\r\nYes, that's true.\r\n \r\n...by the way, the surface area of a cube of a side $s$ is $6s^2$ as you have seen.", "Solution_9": "[quote=\"math92\"]Find the surface area of a cube with a length of 10, and a height of 10. (is that enough information to solve it?)[/quote]\r\n\r\nArea of one side of the cube is $10^2$ or $100$\r\n\r\nSo the entire surface area would be $100*6$ or $600$", "Solution_10": "[hide]The Surface Area of the cube is composed of $6$ squares with Side Lengths $10$. Thus, the Surface Area is $6\\cdot 10^2$ or $600$[/hide]" } { "Tag": [ "calculus", "integration", "function", "limit", "inequalities", "real analysis", "real analysis unsolved" ], "Problem": "Hello, let $ f_n$ a sequence of nonnegative measurable function on a measurable set E, and let the summable function $ f$ be the pointwise limit of the sequence. Suppose that \r\n$ \\lim_{n \\rightarrow \\infty} \\int_E f_n dx \\equal{} \\int_E fdx$\r\nProve that for every measurable subset $ F \\subseteq E$ it is\r\n\r\n$ \\lim_{n \\rightarrow \\infty} \\int_F f_n dx \\equal{} \\int_F f dx$\r\n\r\nMy question:\r\nIs it obvious that $ \\lim_{n \\rightarrow \\infty} \\int_F f_n dx$ exists? BTW I just could prove that\r\n\r\n$ \\int_F fdx \\leq \\liminf_{n \\rightarrow \\infty} \\int_F f_n dx$ using Fatou's Lemma, but how do I prove the opposite inequality?", "Solution_1": "Existence of the limit is a part of what you have to prove. Indeed, by Fatou's lemma $ \\int_F fdx \\leq \\liminf_{n \\rightarrow \\infty} \\int_F f_n dx$. What's more, the same lemma tells you that $ \\int_{E\\setminus F} fdx \\leq \\liminf_{n \\rightarrow \\infty} \\int_{E\\setminus F} f_n dx$. Think about what this last inequality means for the integrals over $ F$.", "Solution_2": "thanks for the hint :)\r\n\r\nit is for all n:\r\n$ \\int_F f_n dx \\equal{} \\int_E f_n dx \\minus{} \\int_{E \\backslash F} f_n dx$\r\nwhich gives\r\n\r\n$ \\limsup \\int_F f_ndx \\leq \\int_E f dx \\plus{} \\limsup \\left ( \\minus{} \\int_{E \\backslash F} f_n dx \\right ) \\leq \\int_E f dx \\minus{} \\liminf \\int_{E \\backslash F} f_n dx \\leq \\int_E f dx \\minus{} \\int_{E \\backslash F} f dx \\equal{} \\int_F f dx$\r\n\r\nHence:\r\n$ \\int_F f dx \\leq \\liminf \\int_F f_n dx \\leq \\limsup \\int_F f_n dx \\leq \\int_F fdx$\r\nso equalities hold and we finish.\r\n\r\nright?", "Solution_3": "Right. One of the inequalities in the chain is actually an identity, since $ \\limsup(\\minus{}\\dots)\\equal{}\\minus{}\\liminf(\\dots)$", "Solution_4": "is it still true if we don't require the $ f_n$s to be nonnegative?\r\nI cannot adapt any counterexample of Fatou's Lemma to disprove this :)", "Solution_5": "Let $ E\\equal{}[\\minus{}1,1]$ and $ f_n(x)\\equal{}\\begin{cases}n,&0