共查询到20条相似文献,搜索用时 15 毫秒
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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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文[1]给出了以下不等式的简证与加强,已知a,b〉0,
(1)求证:√a/2b+a+√b/2a+b≤2/√3
(2)求证:√a/2a+b+√b/2b+a≤2/√3 相似文献
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In this paper, we prove that if a, b and c are pairwise coprime positive integers such that a^2+b^2=c^r,a〉b,a≡3 (mod4),b≡2 (mod4) and c-1 is not a square, thena a^x+b^y=c^z has only the positive integer solution (x, y, z) = (2, 2, r).
Let m and r be positive integers with 2|m and 2 r, define the integers Ur, Vr by (m +√-1)^r=Vr+Ur√-1. If a = |Ur|,b=|Vr|,c = m^2+1 with m ≡ 2 (mod 4),a ≡ 3 (mod 4), and if r 〈 m/√1.5log3(m^2+1)-1, then a^x + b^y = c^z has only the positive integer solution (x,y, z) = (2, 2, r). The argument here is elementary. 相似文献
Let m and r be positive integers with 2|m and 2 r, define the integers Ur, Vr by (m +√-1)^r=Vr+Ur√-1. If a = |Ur|,b=|Vr|,c = m^2+1 with m ≡ 2 (mod 4),a ≡ 3 (mod 4), and if r 〈 m/√1.5log3(m^2+1)-1, then a^x + b^y = c^z has only the positive integer solution (x,y, z) = (2, 2, r). The argument here is elementary. 相似文献
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文[1]证明了这样一个不等式,若n,b,c为正实数,则.√a/b+c+√b/a+c+√c/a+b〉2. 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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题目 已知a1+a2+a3=4,b1+b2+b3=3,且a1,a2,a3,b1,b2,b3均为正数,试求√a1^2+b1^2+√a2^2+b2^2+√a3^2+b3^2的最小值。 相似文献
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题目对任意正实数a、b、c,求证:1〈a/√a^2+b^2+b/√b^2+c^2+c/√c^2+a^2≤3√2/2. 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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In this article, we prove that the double inequality
αP(a,b)+(1-α)Q(a,b)〈M(a,b)〈βP(a,b)+(1-β)Q(a,b)
holds for any a,b 〉 0 with a ≠ b if and only if α≥1/2 and β≤[π(√2 lov (1+√2)-1]/[√2π-2) log (1+√2)]=0.3595…,where M(a, b), Q(a, b), and P(a, b) ave the Neuman-Sandor, quadratic, and first Seiffert means of a and b, respectively. 相似文献
αP(a,b)+(1-α)Q(a,b)〈M(a,b)〈βP(a,b)+(1-β)Q(a,b)
holds for any a,b 〉 0 with a ≠ b if and only if α≥1/2 and β≤[π(√2 lov (1+√2)-1]/[√2π-2) log (1+√2)]=0.3595…,where M(a, b), Q(a, b), and P(a, b) ave the Neuman-Sandor, quadratic, and first Seiffert means of a and b, respectively. 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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文[1]证明了一对有趣的不等式:设a,b,c为正数,且a+b+c=1,则有
(1/b+c-a)(1/c+a-b)(1/a+b-c)≥(7/6)^3,
(1/b+c-a)(1/c+a+b)(1/a+b+c)≥(11/6)^3。
为了推广这两个不等式,文[1]提出下面四个命题,要求证明或否定之. 相似文献