共查询到20条相似文献,搜索用时 312 毫秒
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文[1]给出并证明了如下不等式:
若a,b,c是正数,且a+b+c=1,则有:
(1/b+c -a)(1/c+a -b)(1/a+b -c)≥(7/6)^3 相似文献
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2008年全国高中数学联赛山东赛区预赛第17题:
若x〉0,y〉0,z〉0,且xyz=1,求证:
1〈1/(1+x)+1/(1+y)+1/(1+z)〈2.
原证 (命题组给出的证明)任取a〉0.令b=ax,c=by,由xyz=1,得x=b/a,y=c/b,z=a/c, 相似文献
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不等式中的一对姐妹花 总被引:4,自引:0,他引:4
若a,b,c是正数,且a+b+c=1,则有(1/b+c -a)1/c+a-b)(1/a+b -c)≥(7/6)^3当且仅当a=b=c=1/3时取等号。 相似文献
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对如下一道日本数学奥林匹克试题:
问题1已知a,b,c〉0,求证:(b+c-a)^2/(b=c)^2+a^2+(c+a-b)^2/(c+a)^2+b^2+(a+b-c)^2/(a+b)^2+c^2≥3/5. 相似文献
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贵刊文[1]介绍了俄罗斯杂志《中学数学》刊登的一组不等式,其中之一是下面的瓦西列夫不等式:
设a,b,c〉0,且a+b+c=1,则
a^2+b/b+c+b^2+c/c+a+c^2+a/a+b≥2 (1) 相似文献
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《数学教学》2012年第12期的数学问题874为:题目 已知 m,n∈N+,m,n≥2,xi∈R+(i=1,2,…,m),(^m∑i=1)xi=S,n∈N+,求证:(^m∑i=1)^n√xi/S-xi≥.看完此题,笔者不禁想起了文[1]中的不等式:题源1已知a,b,c为正数,求证:√a/(b+c)+√b/(c+a)+√c/(a+b)〉2。 相似文献
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有这样一道吸引大家眼球的有趣不等式试题:问题1设正实数a,b,c满足abc=1,求证:a2+1(1/2)+b2+1(1/2)+c2+1(1/2)≤2(1/2)(a+b+c)1本刊文[1]通过构造函数f(x)=x2+1(1/2)-2(1/2)x-2(1/2)2lnx(x〉0),借助二阶导数和三元均值不等式给出一个证明.是否有更简单、更初等(即不用导数)的证明呢?笔者经过思考发现,借助平方差公式和二元均值不等式,最终可以获得一个简单、 相似文献
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第20届伊朗数学竞赛中有如下一道三元不等式题:已知a,b,c为正实数,a2+b2+c2+abc=4,求证:a+b+c≤3.如果退化为二元情况,不妨令c=b,则题设条件变为a2+2b2+ab2=4(*),整理得a+b2=2,在此式中再分别令a=x+y/2,b=xy(1/2)或者a=2x+y/3,b=xy(1/2)等,并代入后进行整理,就得到下列几道最值题:问题1已知x, 相似文献
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With the help of continued fractions, we plan to list all the elements of the set Q△ = {aX2 + bXY + cY2 : a,b, c ∈Z, b2 - 4ac = △ with 0 ≤ b 〈 √△}of quasi-reduced quadratic forms of fundamental discriminant △. As a matter of fact, we show that for each reduced quadratic form f = aX2 + bXY + cY2 = (a, b, c) of discriminant △〉0(and of sign σ(f) equal to the sign of a), the quadratic forms associated with f and defined by {〈a+bu+cu2,b+2cu.c〉},with 1≤σ(f)u≤b/2|c| (whenever they exist), 〈c,-b-2cu,a+bu+cu2〉 with b/2|c|≤σ(f)u≤[w(f)]=[b+√△/2|c|], are all different from one another and build a set I(f) whose cardinality is #I(f)={1+[ω(f)],when(2c)|b,[ω(f)],when (2c)|b. If f and g are two different reduced quadratic forms, we show that I(f) ∩ I(g) = Ф. Our main result is that the set Q△ is given by the disjoint union of all I(f) with f running through the set of reduced quadratic forms of discriminant △〉0. This allows us to deduce a formula for #(Q△) involving sums of partial quotients of certain continued fractions. 相似文献
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瓦西列夫不等式:
设n,b,c〉0,n+b+c=1,则a^2+b/b+c+b^2+c/c+a+c^2+a/a+b≥2. 相似文献
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Jun-ping Li 《应用数学学报(英文版)》2006,22(4):663-670
Suppose {X(t); t≥ 0} is a single birth process with birth rate qii+l (i 〉 0) and death rate qij (i 〉 j ≥ 0). It is proved in this paper that (i) if there exists aconstant c≥ 0 such that b(i)-a(i)+ci is nondecreasing with respect to i and a(i) + u(i) - ci ≥ 0 (i≥ 0), then
VarX(t)-EX(t)≥-X(0)e^-2ct,t≥0,
or (ii) if there exists a constant u(i) - c≥ 0 such that b(i)-a(i)+ci is non-increasing with respect to i and a(i)+u(i)-ci≤0(i≥0),then
VarX(t) - EX(t) ≤ -X(0)e^-2c,t ≥ 0
Hereb(i) = qii+1, a(0) = 0, a(i) = ∑j=^ijqii-j (i≥ 1), u(0) = u(1) =0 and u(i) = 1/2∑j=^ij(j - 1)qii-j (i ≥ 2) . This result covers the results for birth-death processes obtained in [7]. 相似文献
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瓦西列夫不等式的加强 总被引:2,自引:0,他引:2
本刊曾刊登了瓦西列夫提出的如下优美的不等式:设a,b,C〉0,a+b+c=1,则,^2a+b/b+c+b^2+c/c+a+c^2+a/+a+b≥2①笔者经过探索,得到了①的一个加强结果: 相似文献