[bdt] bĐt muirhead
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hgjhgjghjTRANSCRIPT
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1BT NG THC MUIRHEAD V MT VI P DNG
L H QU, Trng THPT Duy Tn, Kon Tum
Bt ng thc Muirhead l mt dng tng qut rt quan trng ca bt ngthc AM-GM. N l mt cng c rt mnh trong vic gii mt s bi ton v bt ngthc.
1. nh l Muirhead
1.1. nh ngha 1 (B tri)
Cho hai b s thc bt k a = (a1, a2, ..., an) v b = (b1, b2, ..., bn). Ta ni b a tri hnb b, k hiu a b nu chng tha mn cc iu kin sau y
i) a1 a2 ... an v b1 b2 ... bn,ii) a1 b1, a1 + a2 b1 + b2, ..., a1 + a2 + ...+ an b1 + b2 + ...+ bn viii) a1 + a2 + ...+ an = b1 + b2 + ...+ bn.
1.2. nh ngha 2 (Trung bnh loi [a])
Gi s xi > 0, 1 i n. K hiu!F (x1, x2, ...xn)
l tng gm n! biu thc thu c t F (x1, x2, ...xn) bng tt c cc hon v ca xi. Tas ch xt trng hp c bit
F (x1, x2, ..., xn) = x11 x
22 ...x
nn , vi xi > 0, ai > 0.
Khi trung bnh loi [a] c nh ngha bi
[a] = [a1, a2,..., an] =1
n!
!x11 x
22 ...x
nn .
c bit
[1, 0, 0,..., 0] = (n1)!n!
(x1 + x2 + ...+ xn) =1n
ni=1
xi l trung bnh cng ca xi.[1n, 1n,..., 1
n
]=n!
n!
(x
1n1 .x
1n1 ...x
1n1
)= nx1x2...xn l trung bnh nhn ca xi.
Khi a1 + a2 + ...+ an = 1 th [a] l m rng thng thng ca trung bnh cng v trungbnh nhn.
1.3. nh ngha 3
Gi P (x, y, z) l mt hm ba bin x, y, z. Khi , ta nh nghai) Tng hon v:
cyclic
P (x, y, x) = P (x, y, z) + P (y, z, x) + P (z, x, y).
ii) Tng i xng:sym
P (x, y, x) = P (x, y, z) + P (x, z, y) + P (y, x, z) + P (y, z, x)+P (z, x, y)+P (z, y, x).
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21.4. nh l Muirhead
nh l 1 (Bt ng thc Muirhead). Cho xi > 0, 1 i n v a, b l hai b n sthc. Nu a b th [a] [b].ng thc xy ra khi v ch khi a 6= b v x1 = x2 = ... = xn.Chng minh. C th xem phn chng minh nh l Muirhead trong cc ti liu thamkho [1], [2], [3], [4].V rng (1, 0, ..., 0) ( 1
n, 1n, ..., 1
n
), nn bt ng thc AM-GM l mt h qu ca bt
ng thc Muirhead.
2. Mt vi p dng
phn tip theo, chng ti xin trnh by mt s p dng ca bt ng thcMuirhead trong vic chng minh bt ng thc.
2.1 Chng minh cc bt ng thc i s
V d 1. Cho ba s thc dng a, b, c. Chng minh rng
(a+ b)(b+ c)(c+ a) 8abc.
Li gii. Khai trin v rt gn ta c bt ng thc tng ng
a2b+ a2c+ b2c+ b2a+ c2a+ c2b 6abc.
V (2, 1, 0) (1, 1, 1) nn theo bt ng thc Muirhead ta c [(2, 1, 0)] [(1, 1, 1)].ng thc xy ra khi v ch khi a = b = c.
V d 2 (Yogoslavia-1991). Chng minh rng vi mi s thc dng a, b, c, ta lun c
1
a3 + b3 + abc+
1
b3 + c3 + abc+
1
c3 + a3 + abc 1abc
.
Li gii. Quy ng v b mu, ri nhn hai v cho 2, ta c bt ng thc tng ngsym
(a3 + b3 + abc)(b3 + c3 + abc) 2(a3 + b3 + abc)(b3 + c3 + abc)(c3 + a3 + abc)
sym
(a7bc+ 3a4b4c+ 4a5b2c2 + a3b3c3) sym
(a3b3c3 + 2a6b3 + 3a4b4c+ a7bc+ 2a5b2c2)
sym
(2a6b3 2a5b2c) 0
V (6, 3, 0) (5, 2, 2) nn theo bt ng thc Muirhead nn v tri ca bt ng thccui cng l mt hng t khng m. T , ta c iu phi chng minh.
Nhn xt 1. v d tip theo, chng ta s s dng n mt k thut rt hu ch sauy
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3Khi x1.x2...xn = 1 th [(a1, a2, ..., an)] = [(a1 r, a2 r, ..., an r)], vi r R.
V d 3 (IMO-1995). Cho a, b, c l cc s thc dng tha mn iu kin abc = 1.Chng minh rng
1
a3(b+ c)+
1
b3(c+ a)+
1
c3(a+ b) 3
2.
Li gii 1. Quy ng v b mu, ta c bt ng thc tng ng
2(a4b4 + b4c4 + c4a4) + 2(a4b3c+ a4c3b+ b4a3c+ c4a3b+ c4b3a) + 2(a3b3c2 + b3c3a2 + c3a3b2)
3(a5b4c3 + a5c4b3 + b5c4a3 + b5a4c3 + c5a4b3 + c5b4a3) + 6a4b4c4.
S dng k hiu [a], ta c bt ng thc tng ng
[(4, 4, 0)] + 2[(4, 3, 1)] + [(3, 3, 2)] 3[(5, 4, 3)] + [(4, 4, 4)]
rng 4 + 4 + 0 = 4 + 3 + 1 = 3 + 3 + 2 = 8, nhng 5 + 4 + 3 = 4 + 4 + 4 = 12.Bi vy, ta c th chn r = 4
3v s dng k thut trn ta c [(5, 4, 3)] =
[(113, 83, 53)].
Hn na [(4, 4, 4)] =[(83, 83, 83)].
p dng bt ng thc Muirhead cho ba b s (4, 4, 0) (113, 83, 53
), (4, 3, 1) (11
3, 83, 53
),
(3, 3, 2) (83, 83, 83
)v cng cc bt ng thc va nhn c ta c bt ng thc phi
chng minh.
Li gii 2. Bt ng thc cho tng ng vi
1
a3(b+ c)+
1
b3(c+ a)+
1
c3(a+ b) 3
2(abc)43
t a = x3, b = y3, c = z3, vi x, y, z > 0. Khi bt ng thc tr thnhcyclic
1
x9(y3 + z3) 3
2x4y4z4.
Quy ng mu s chung v b mu, ta c bt ng thcsym
x12y12 + 2sym
x12y9z3 +sym
x9y9z6 3sym
x11y8z5 +sym
x8y8z8
hay(sym
x12y12 sym
x11y8z5)+2(sym
x12y9z3sym
x11y8z5)+(sym
x9y9z6sym
x8y8z8) 0.
V (12, 12, 0) (11, 8, 5), (12, 9, 3) (11, 8, 5), (9, 9, 6) (8, 8, 8) nn theo bt ng thcMuirhead th mi hng t trong bt ng thc cui l khng m. T ta c iu phichng minh.
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4V d 4 (IMO Shortlist-1998). Cho cc s thc dng x, y, z tha mn iu kinxyz = 1. Chng minh rng
x3
(1 + y)(1 + z)+
y3
(1 + z)(1 + x)+
z3
(1 + x)(1 + y) 3
4.
Li gii. Quy ng mu s chung v b mu, ta c bt ng thc tng ng
4(x4 + y4 + z4 + x3 + y3 + z) 3(1 + x+ y + z + xy + yz + zx+ xyz).
S dng k hiu [a], ta c bt ng thc tng ng
4[(4, 0, 0)] + 4[(3, 0, 0)] [(0, 0, 0)] + 3[(1, 0, 0)] + 3[1, 1, 0] + [(1, 1, 1)].
p dng bt ng thc Muirhead v k thut trn, ta c
[(4, 0, 0)] [(43,4
3,4
3
)]= [(0, 0, 0)]
3[(4, 0, 0)] 3[(2, 1, 1)] = 3[(1, 0, 0)]3[(3, 0, 0)] [(4
3,4
3,1
3
)]= 3[(1, 1, 0)]
v[(3, 0, 0)] [(1, 1, 1)].
Cng cc bt ng thc va nhn c ta c bt ng thc phi chng minh. ng thcxy ra khi v ch khi x = y = z = 1.
Nhn xt 2.i) Trong bi ton trn, ta c th "ni" bt iu kin rng buc thnh iu kin rng
hn xyz 1. Khi , ta c bt ng thc
[(a1, a2, a3)] [(a1 r, a2 r, a3 r)],
trong r 0 bt k.ii) S dng bt ng thc AM-GM, ta d dng chng minh c kt qu sau:Vi hai b n s thc a v b, ta lun c
[a] + [b]
2 [a+ b
2
].
V d 5 (IMO-2005). Cho x, y, z l cc s thc dng sao cho xyz 1. Chng minhrng
x5 x2x5 + y2 + z2
+y5 y2
y5 + z2 + x2+
z5 z2z5 + x2 + y2
0.
Li gii. Sau khi quy ng mu s chung, b mu v s dng k hiu [a], ta c btng thc tng ng
[(9, 0, 0)]+4[7, 5, 0]+[(5, 2, 2)]+[(5, 5, 5)] [(6, 0, 0)+[(5, 5, 2)]+2[(5, 4, 0)]+2[(4, 2, 0)]+[(2, 2, 2)].
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5Ta c(1) [9, 0, 0] [(7, 1, 1)] [(6, 0, 0)] (do bt ng thc Muirhead v nhn xt 2.i))(2) [(7, 5, 0)] [(5, 5, 2)] (do bt ng thc Muirhead)(3) 2[(7, 5, 0)] 2[(6, 5, 1)] 2[(5, 4, 0)] (do bt ng thc Muirhead v nhn xt 2.i))(4) [(7, 5, 0)] + [(5, 2, 2)] 2[(6, 7
2, 1)] 2[(11
2, 72, 32
)] 2[(4, 2, 0)](do bt ng thc Muirhead v nhn xt 2)(5) [(5, 5, 5)] [(2, 2, 2)] (do nhn xt 2.i))
Cng cc bt ng thc trn v vi v, ta c bt ng thc cn chng minh.
2.2 Chng minh cc bt ng thc hnh hc
V d 6 (IMO-1961). Cho a, b, c l di ba cnh ca tam gic ABC, S l din tchca tam gic . Chng minh rng
43S a2 + b2 + c2.
Li gii 1. S dng cng thc Heron, ta c th vit li bt ng thc cho nh sau
a2 + b2 + c2 43
(a+ b+ c)
2
(a+ b c)2
(a+ c b)2
(b+ c a)2
.
Bnh phng c hai v ca bt ng thc, ta c
(a2 + b2 + c2)2 3[(a+ b)2 c2)(c2 (b a)2)] == 3(2c2a2 + 2c2b2 + 2a2b2 (a4 + b4 + c4))
a4 + b4 + c4 a2b2 + b2c2 + c2a2
Dng k hiu [a], ta c bt ng thc tng ng
[(4, 0, 0)] [(2, 2, 0)]Bt ng thc cui cng ny lun ng do bt ng thc Muirhead ng vi hai b s(4, 0, 0) (2, 2, 0).
Li gii 2. tx = a+ b c, y = c+ a b, z = b+ c a,
ta thu c x+ y + z = a+ b+ c, khi , dng cng thc Heron ta c
4S =(a+ b+ c)(xyz)
(a+ b+ c)
(x+ y + z)3
27=
(a+ b+ c)2
33
.
Lc ny, ta ch cn chng minh (a + b + c)2 3(a2 + b2 + c2). Bt ng thc cui cngny c suy ra t bt ng thc Muirhead, v rng [(1, 1, 0)] [(2, 0, 0)].
V d 7. Xt tam gic ABC vi di cc cnh l a, b, c v R, r ln lt l bn knh cang trn ngoi tip, ni tip ca tam gic . Chng minh rng
r
R2[2a2 (b c)2][2b2 (c a)2][2c2 (a b)2]
(a+ b)(b+ c)(c+ a).
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6Li gii. Trc ht, ta thun nht bt ng thc phi chng minh vi cc bin x, y, zbng cch dng cc ng nht thc
R =abc
4S, r =
S
p, S2 = p(p a)(p b)(p c), p = a+ b+ c
2
v t
a =y + z
2, b =
z + x
2, c =
x+ y
2.
Khi , ta c
x = b+ c a > 0, y = c+ a b > 0, z = a+ b c > 0.Bnh phng hai v v rt gn ta c bt ng thc tng ng
105[(4, 4, 4)] + 264[(5, 4, 3)] + 88[(6, 3, 3)] + 48[(7, 3, 2)] + 9[(8, 2, 2)] 136[(5, 5, 2)] + 106[(6, 4, 2)] + 176[(6, 5, 1)] + 7[(6, 6, 0)] + 72[(7, 4, 1)]++ 8[(7, 5, 0)] + 8[(8, 3, 1)] + [(8, 4, 0)].
p dng bt ng thc Muirhead, ta ln lt c
9[(8, 2, 2)] 8[(8, 3, 1)] + [(8, 4, 0)]48[(7, 3, 2)] 48[(7, 4, 1)]88[(6, 3, 3)] 88[(6, 5, 1)]264[(5, 4, 3)] 136[(5, 5, 2)] + 106[(6, 4, 2)] + 22[(6, 5, 1)]105[(4, 4, 4)] 66[(6, 5, 1)] + 7[(6, 6, 0)] + 24[(7, 4, 1)] + 8[(7, 5, 0)].
Cng cc v tng ng ca cc bt ng thc trn, ta c bt ng thc cn chngminh. ng thc xy ra khi v ch khi x = y = z hay tam gic ABC u.
Bi tp
Bi 1. Chng minh rng mi s thc dng a, b, c, ta lun c
a5 + b5 + c5 a3bc+ b3ca+ c3ab.Bi 2. Cho a, b, c l cc s thc dng. Chng minh rng
a
(a+ b)(a+ c)+
b
(b+ c)(b+ a)+
c
(c+ a)(c+ b) 9
4(a+ b+ c).
Bi 3 (IMO-1964). Cho a, b, c l cc s thc dng. Chng minh rng
a3 + b3 + c3 + 3abc ab(a+ b) + bc(b+ c) + ca(c+ a).Bi 4 (Iberoamerican Shortlist-2003). Cho ba s thc dng a, b, c. Chng minhrng
a3
b2 bc+ c2 +b3
c2 ca+ a2 +c3
a2 ab+ b2 a+ b+ c.
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7Bi 5 (IMO-1984). Cho x, y, z l cc s thc khng m tha mn iu kin x+y+z = 1.Chng minh rng
0 xy + yz + zx 2xyz 727.
Bi 6. Cho x, y, z l cc s thc khng m sao cho xy + yz + zx = 1. Chng minh
1
x+ y+
1
y + z+
1
z + x 5
2.
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8Ti liu tham kho
[1] G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Cambridge University Pree,1967.
[2] Radmila Bulajich Manfrino, Jos Antonio Gosmez Ortega, Rogelio valdez Delgado,Inequalities, A Mathematical Olympial Approach, Birkhauser, 2009.
[3] Lau Chi Hin, Muirheads Inequality, Vo.11. Mathematical Excalibur, 2006.
[4] Ivan Mati, Classical Inequalities, Olympial Training Materials, 2007.
[5] Zoran Kadelburg, Dusan uki, Milivoje Luki, Ivan Mati, Inequalities of Karamata,Schur and Muirhead, and some applications, The teaching of Mathematics, Vol. VIII,pp. 31-45, 2005.
[6] Cezar Lupu, Tudorel Lupu, Problem 11245, American mathematical Monthly,Vol.113, 2006.
[7] Andre Rzym, Muirheads Inequality, November 2005.
[8] Stanley Rabinowitz, On The Computer Solution of Symmetric Homogeneous TriangleInequalities, Alliant Computer Systems Corporation Littleton, MA 01460.
[9] Nguyn Vn Mu, Bt ng thc, nh l v p dng, NXB Gio Dc, 2005.
[10] Phm Kim Hng, Sng to bt ng thc, NXB H Ni, 2007.
[11] Ng Th Phit, Mt s phng php mi trong chng minh bt ng thc, NXB GioDc, 2007.
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