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S GIO DC V O TO KHNH HATRNG THPT CHUYN L QU N
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Chuyn :
BT NG THC C SI
TRONG CC K THI TUYN SINHI HC V CAO NG
Gio vin : Hunh Kim Linh
Nm hc 2008 - 2009
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Li ni u :Thc hin nhim v nm hc 2008 2009, Trng THPT Chuyn L
Qu n Khnh Ha c khuyn khch cc gio vin dy mn chuyn lmchuyn xy dng ti nguyn ca t chuyn mn
Chnh v vy ti thc hin v lm chuyn v
BT NG THC C SITRONG CC K THI TUYN SINH I HC V CAO NG
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BT NG THC C SITRONG CC K THI TUYN SINH I HC V CAO NG
Hunh Kim Linh
Trong cc k thi tuyn sinh i hc v cao ng, c mt hay hai cu kh phn loi th sinh v thng c mt cu v bt ng thc.
1) nh l ( Bt ng thc C si):Cho n s thc khng m :1 2 na ,a ,..., a .Ta c :
1 2 n n1 2 n
a a ... a a a ...an
+ + + .
ng thc xy ra khi v ch khi1 2 na a a= = =L .
2) Mt s bt ng thc lin quan n bt ng thc C si :2.1) Cc Bt ng thc dng phn thc
Vi x, y > 0. Ta c :
( )1 1 4 1x y x y+ +
( )( )2
1 4 2xy x y
+
ng thc xy ra khi v ch khi x = y.
Chng minh : (1) v( ) ( )22 x y 4xy + (ng)
Vi x, y, z > 0. Ta c :
( )1 1 1 9 3x y z x y z+ + + +ng thc xy ra khi v ch khi x = y = z.
2.2) Cc bt ng thc dng a thc :( )2 2 2x y z xy yz zx 4+ + + +
( ) ( ) ( )22 2 23 x y z x y z 5+ + + +
( ) ( ) ( )2x y z 3 xy yz zx 6+ + + +
ng thc xy ra khi v ch khi x = y = z.3
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3) MT S BI TON THI I HC :Bi ton 1 : thi tuyn sinh i hc khi A nm 2005
Cho x, y, z l cc s dng tha mn :1 1 1 4x y z
+ + = .
Chng minh rng :1 1 1 1.
2x y z x 2y z x y 2z+ + + + + + + +
Li gii :Cch 1 :
p dng bt ng thc :1 1 4x y x y
+ + vi x, y > 0, ta c :
( )1 1 1 1 1 1 1 1 1 1 1 18 2 4 1x y z x y y z z x x y y z z x
= + + = + + + + + + + + + + Tng t
( )
1 1 1 1 1 1 1 1 12x y y z z x x y x z x y y z y z z x
1 1 14 22x y z x 2y z x y 2z
+ + = + + + + + + + + + + + + + + + + + + + +
T (1) v (2) suy ra
1 1 18 82x y z x 2y z x y 2z + + + + + + + +
1 1 1 1.2x y z x 2y z x y 2z
+ + + + + + + +
ng thc xy ra khi 3x y z .4= = =
Cch 2 :
p dng bt ng thc :1 1 4x y x y
+ + vi x, y > 0, v bt ng thc Csi ta c :
( ) ( ) ( )2x y z x y x z 2 xy xz+ + = + + + +Do :
1 1 1 1 1 12x y z 2 8xy xz xy xz
+ + + + Tng t :
1 1 1 1x 2y z 8 xy yz
+ + +
1 1 1 1x y 2z 8 xz yz
+ + +
Cng v theo v 3 bt ng thc trn ta c :
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( )1 1 1 1 1 1 1 32x y z x 2y z x y 2z 4 xy yz zx
+ + + + + + + + + +
Mt khc theo bt ng thc Csi
( )1 1 1 1 1 1 1 1 1 1 1 14 42 x y 2 y z 2 z x xy yz zx
= + + + + + + +
T (3) v (4) suy ra :1 1 1 1.
2x y z x 2y z x y 2z+ + + + + + + +
Cch 3 :
p dng bt ng thc Csi cho 4 s dng
( ) 1 1 1 1x x y z 16x x y z + + + + + +
Suy ra1 1 2 1 1
2x y z 16 x y z + + + +
Tng t1 1 1 2 1
x 2y z 16 x y z + + + +
1 1 1 1 2x y 2z 16 x y z
+ + + + Cng v theo v 3 bt ng thc trn ta c :
1 1 1 1.2x y z x 2y z x y 2z
+ + + + + + + +
M rng bi ton 1 :Cho n s thc dng cho trc :1 2 na ,a , aK tha iu kin :
1 2 n
1 1 1ka a a+ + + =L . Vi n > 1 v k > 0 cho trc. Chng minh rng :
+ + + + + + + + + + + + + + +
LL L L L1 1 2 2 n n 2 1 n n 1 1 n n 1 1 2 n 1 n 1 2 n
1 1 1 kma m a m a m a m a ma m a ma m a m m m
Bi ton 2 : thi tuyn sinh i hc khi B nm 2005Chng minh rng : vi mi xR, ta c :
x x xx x x12 15 20
3 4 55 4 3
+ + + + Khi no ng thc xy ra.
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Li p dng bt ng thc Csi
( )2
1 1 1 13 3 2xy yz zx xyz
+ + =
T (1) v (2) suy ra iu cn chng minh.ng thc xy ra khi v ch khi x = y = z = 1.
M rng b i ton 3 :Cho cc s thc dng1 2 na ,a , aK tha mn : 1 2 na .a a 1=L
Chng minh rng :
( ) ( ) ( )
p p p p p p pm m m1 n 1 2 n n 1 n 2 m
q q q1 2 n 1 2 3 n n 1 n 2
1 a a 1 a a 1 a a an n
aa a a a a a a a
+ + + + + + ++ + + L L LLL L L
Trong m 2 l s nguyn dng, p, q l cc s thc ty
Hng dn :p dng bt ng thc Csi cho n s
( )pp p n
1 n 1 1 2 n 11 a a n a .a a + + L L
Bi ton 4 :D B 1 KHI A Nm 2005 :Cho x, y, z l ba s tha x + y + z = 0. Chng minh rng :
3 4 3 4 3 4 6 x y z + + + + +
Li gii :Ta c:
4x x x3 4 1 1 1 4 4 4+ = + + + + = 84x x x3 4 2 4 2. 4 .
Tng t
+ 8y x3 4 2 4 v 8z z3 4 2 4+
Vy+ + + + + + + + + =
38 8 8 8 24x y z x y z x y z x y z3 4 3 4 3 4 2 4 4 4 6 4 .4 .4 6 4 6
ng thc xy ra khi v ch khi x = y = z = 0.
Bi ton 5 :D B 2 KHI A Nm 2005 :
Chng minh rng : vi mi x, y > 0 ta c :( )2
91 1 1 256 + + +
y x
x y
.
ng thc xy ra khi no?Li gii :
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Ta c:3
43
x x x x1 x 1 4
3 3 3 3+ = + + + . ng thc xy ra khi v ch khi x = 3
34
3 3y y y y y
1 1 4x 3x 3x 3x 3 .x
+ = + + + . ng thc xy ra khi v ch khi y = 3x = 9
( )
3
4 39 3 3 3 3
1 1 4y y y y y
+ = + + + 2 6
43
9 31 16
y y
+
.
ng thc xy ra khi v ch khi y = 9
Vy ( ) + + + =
2 3 3 64
3 3 3 3y 9 x y 3
1 x 1 1 256 256x y 3 3 .x y
ng thc xy ra khi v ch khi x = 3 v y = 9.
Bi ton 6 :D B 1 KHI B Nm 2005 :
Cho a, b, c l ba s dng tha mn : a + b + c =34 . Chng minh rng :
3 3 33 3 3 3a b b c c a+ + + + + .Khi no ng thc xy ra ?
Li gii :Cch 1:
Ta c :
( ) ( )
( ) ( )
( ) ( )
3
3
3
a 3b 1 1 1a 3b 1.1 a 3b 2
3 3b 3c 1 1 1
b 3c 1.1 b 3c 23 3
c 3a 1 1 1c 3a 1.1 c 3a 23 3
+ + ++ = + +
+ + ++ = + +
+ + ++ = + +
Suy ra
( ) + + + + + + + + + = 3 3 3 1 1 3a 3b b 3c c 3a 4 a b c 6 4. 6 3
3 3 4
Du = xy ra3
a b c 1a b c4
4a 3b b 3c c 3a 1
+ + = = = =
+ = + = + =Cch 2:
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t33x a 3b x a 3b= + = +
= + = +33y b 3c y b 3c
= + = +33z c 3a z c 3a
( )3 3 3 3
x y z 4 a b c 4. 34+ + = + + = =.Bt ng thc cn chng minhx y z 3 + + .Ta c :
33 3x 1 1 3 x .1.1 3x+ + =3 33y 1 1 3 y .1.1 3y+ + =
33 3
z 1 1 3 z .1.1 3z+ + = ( )9 3 x y z + + (V 3 3 3x y z 3+ + = ).
Vyx y z 3+ +
Hay3 3 3a 3b b 3c c 3a 3+ + + + +
ng thc xy ra khi v ch khi= = =1a b c 4
Bi ton 7 :D B 2 KHI B Nm 2005 :Chng minh rng nu0 1 y x th
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x y y x .
ng thc xy ra khi no?Li gii :
Ta c20 x 1 x x
1 1x y y x x y y x
4 4 + (1)
Theo bt ng thc Cauchy :
+ + =2 21 1 1y x yx 2 yx . x y4 4 4 1x y y x 4
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Du = xy ra =
= = =
2
2
0 y x 1 x 1x x 1
y1 4yx4
Bi ton 8 :D B 1 KHI D Nm 2005 :Cho x, y, z l ba s dng v x yz = 1. Chng minh rng :
2 2 2 31 1 1 2 x y z
y z x+ + + + + .
Li gii : Ta c:
2 2x 1 y x 1 y2 . x1 y 4 1 y 4
+ ++ =+ +
2 2y 1 z y 1 z2 y
1 z 4 1 z 4+ ++ =+ +
2 2z 1 x z 1 x2 z
1 x 4 1 x 4+ ++ =+ +
Cng ba bt ng thc trn v theo v ta c:
( )2 2 2x 1 y y 1 z z 1 x x y z
1 y 4 1 z 4 1 x 4 + + ++ + + + + + + + + +
( ) ( )+ ++ +
+ + + + + + + +2 2 2 3 x y zx y z 3 x y z 3
x y z1 y 1 z 1 x 4 4 4 4
3 3 9 3 6 3.34 4 4 4 4 2
= = =( v 3x y z 3 xyz 3+ + =)
Vy2 2 2x y z 3
1 y 1 z 1 x 2+ + + + +
Bi ton 9 : thi tuyn sinh i hc khi A nm 2006
Cho hai s thcx 0, y 0 thay i v tha mn iu kin:( ) 2 2x y xy x y xy+ = + .
Tm gi tr ln nht ca biu thc 3 31 1A .
x y
= +
Li gii :
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T gi thit suy ra: 2 21 1 1 1 1 .x y x y xy+ = +
t1 1a, bx y
= = ta c: ( )2 2a b a b ab 1+ = +
Khi ( )( ) ( )23 3 2 2A a b a b a b ab a b .= + = + + = +
T (1) suy ra: ( )2
a b a b 3ab.+ = +
V2
a bab
2+
nn ( ) ( )2 23a b a b a b4+ + +
( ) ( )2a b 4 a b 0 0 a b 4 + + +
Suy ra: ( )2A a b 16.= +
Vi 1x y2
= = th A 16.= Vy gi tr ln nht caA l 16.
Bi ton 10 : D b 1 i hc khi A nm 2006Cho cc s thcx,y,z tha mn iu kin:x y z3 3 3 1. + + = Chng minh rng:
x y z x y z
x y z y z x z x y
9 9 9 3 3 3 .43 3 3 3 3 3+ + +
+ ++ + + + +
Li gii : t x y z3 a,3 b,3 c= = = . Ta c:a,b,c 0> v 1 1 1 1 ab bc ca abc.a b c+ + = + + =
Bt ng thc cn chng minh tng ng vi:2 2 2a b c a b c
a bc b ca c ab 4+ ++ + + + +
3 3 3
2 2 2
a b c a b ca abc b abc c abc 4
+ + + +
+ + +
( ) ( ) ( ) ( ) ( ) ( )3 3 3a b c a b c (1).
a b a c b c b a c a c b 4+ + + + + + + + + +
p dng bt ng thc C si ta c
( ) ( ) ( ) ( )
3 3
3a a b a c a a b a c 33 . . a (2)
a b a c 8 8 a b a c 8 8 4+ + + ++ + =+ + + +
( ) ( ) ( ) ( )
3 3
3 b b c b a b b c b a 3
3 . . b (3) b c b a 8 8 b c b a 8 8 4
+ + + ++ + =+ + + +
( ) ( ) ( ) ( )
3 3
3c c a c b c c a c b 33 . . c (4).
c a c b 8 8 c a c b 8 8 4+ + + ++ + =+ + + +
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Cng theo tng v cc bt ng thc (2), (3), (4) ta suy ra
( ) ( ) ( ) ( ) ( ) ( )
3 3 3a b c a b c.a b a c b c b a c a c b 4
+ ++ + + + + + + +
Vy (1) ng v ta c iu phi chng minh.
Bi ton 11 : D b 1 i hc khi B nm 2006Tm gi tr nh nht ca hm s:
211 7
y x 4 1 ,2x x
= + + + vi x 0> .
Li gii :
p dng bt ng thc :( ) ( )( )22 2 2 2a b c d ac bd+ + +
Ta c : ( )
2
2
7 79 7 1 3x x
+ + +
11 1 7y x 32x 2 x
+ + + 9 3 3 15x 6x 2 2 2
= + + + =
Khi x = 3 th y =152 nn gi tr nh nht ca y l152 .
Bi ton 12 : D b 2 i hc khi B nm 2006
Cho hai s dng x, y thay i tha mn iu kinx y 4.+ Tm gi tr nh nht ca biu thc :
2 3
2
3x 4 2 yA .4x y
+ += +
Li gii :
Ta c A =2 3
2 2
3x 4 2 y 3x 1 2 y4x 4 xy y
+ ++ = + + +
A 2x 1 1 y y x y24 x 8 8 2y
+= + + + + +
3 91 2 .2 2
+ + =
Vi x = y = 2 th A =92 .
Vy gi tr nh nht ca A l92
Bi ton 13 : D b i hc khi A nm 2007Cho x, y, z l cc bin s dng. Tm gi tr nh nht ca biu thc:
3 3 3 3 3 33 332 2 2
x y zP 4(x y ) 4(x z ) 4(z x ) 2
y z x
= + + + + + + + +
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Li gii : Vi x, y > 0 ta chng minh :
4(x3 + y3) (x + y)3 () Du = xy rax = yTht vy () 4(x + y)(x2 xy + y2) (x + y)3
4(x2 xy + y2) (x + y)2 do x, y > 0
3(x2
+ y2
2xy) 0 (x y)2
0 (ng)Tng t ta c
4(y3 + z3) (y + z)3 Du = xy ray = z4(z3 + x3) (z + x)3 Du = xy raz = x
Do
( ) ( ) ( )( )3 3 3 3 3 3 33 3 34 x y 4 y z 4 z x 2 x y z 6 xyz+ + + + + + +
Ta li c
3222 xyz6
xz
zy
yx2
++ Du = xy rax = y = z
Suy ra
12xyz1xyz6P
33
+
Du = xy ra
==
=
zyx
1x y z x = y = z = 1
Vy minP = 12 khi x = y = z = 1
Bi ton 14 : D b i hc khi D nm 2007Cho a, b > 0 tha mnab + a + b = 3.Chng minh :
23
baba
ab1a
b31b
a3 22 +++++++ .
Li gii : T gi thita, b > 0 v ab + a + b = 3. Suy ra:
ab 3 (a b)= + , (a+1)(b+1) = ab +a +b + 1 = 4Bt ng thc cho tng ng vi
2 2 3 3a(a 1) 3b(b 1) 3a b 12 (a 1)(b 1) a b
+ + ++ + + + + +
( )( ) 1
ba
3ba
4
3ba
4
3
2
3ba 2222
+++++++
( ) ( ) ( ) 4ba
12ba3ba36ba4 2222
+++++++
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( )2 2 12a b 3 a b 10a b + + + + (*)
t 2 2x a b 0 x (a b) 4ab 4(3 x)= + > = + = 2x 4x 12 0 x 6hay x 2 + x 2 ( v x > 0)
2 2 2x a b 2ab= + + 2 2 2 2a b x 2(3 x) x 2x 6 + = = +
Khi bt ng thc (*) thnh2 12x x 4 0
x + , vi x 2
3 2x x 4x 12 0 + , vi x 2 ( )( )2x 2 x x 6 0 + + , vi x 2 (hin nhin ng)Vy bt ng thc cho c chng minh.
4) Mt s bi ton cc bn t lm :Bi ton 1 5 : Cho x, y > 0 tha : x + y +z = 1 . Chng minh : xyz y x 16+Bi ton 16 : Chng minh rng vi a + b + c = 0 th cbacba 222888 ++++
Bi ton 17 : Cho a, b, c >0 : a + b + c = 1 . Chng minh(1 + a)(1 + b)(1 + c) 8(1 a)(1 b)(1 c)
Bi ton 18 : Cho a, b, c > 0. Chng minh :
bac
acb
cba
bac
acb
cba
+++++ 0 tha : 21 11 11 1 +++++ cba . Chng minh : 81abc
Bi ton 20 : Chng minh rng : vi a> b 0 th ( )( )3
14
2 ++
bbaa
Bi ton 21 : Chng minh : ( )333444
21
cbaba
cac
bcb
a +++++++vi a,b,c > 0
Bi ton 22 : Cho cc s thc dnga,b,c,d tha : 13333 =+++ d cba . Chng minh :
3443
333
2
333
2
333
2
333
2
++
+++
+++
+++ bca
d d ba
cd ca
bd cb
a
Bi ton 23 : Cho cc s thc dnga,b,c,d
tha :a + b + c + d = 1 . Chng minh :( ) ( ) ( ) ( )
2562333
12333
12333
12333
12222 ++
+++
+++
+++
=abcd d bacd acbd bca
P
Bi ton 24 : Cho a. b. c > 0. Chng minh : 3 abccba
ac
cb
ba ++++
Bi ton 25 : Cho a, b, c, d > 0. Chng minh :d cbad cba ++++++
6416411
Bi ton 26 : Cho a, b, c > 0 Chng minh :
cbacbacbacba 24
24
24111
++++++++++Bi ton 27 : Cho x, y, z l cc s thc dng tha tch xyz = 1. Chng minh:
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( ) ( ) ( ) ( ) ( ) ( )
3 3 3 31 1 1 1 1 1 4
x y z y z z x x y
+ + + + + + + +
Bi ton 28 : Cho a, b, c > 0 Chng minh :2111
34111
+++++++ accbbacabcab
Bi ton 29 : Cho a.b > 0 v 1+ba . Chng minh : 1432 22 ++
baabBi ton 30 : Cho a, b, c > 0. Chng minh :
( )( ) ( )( ) ( )( )( )23
221
221
221
cbaacabcacbbcba ++++++++++
Bi ton 31 : Cho a, b, c > 0 tha a + b + c = 1 . Chng minh
92
12
12
1222 +
++
++ abccabbca
Bi to n 32 : Cho x, y, z > 0 tha : x + y +z = 1. Chng minh :
43
111 +++++ z z
y y
x x
Bi ton 33 : Chng minh vi a,b,c > 0 th : ++++ cbacbacba 111333888
Bi ton 34 : Cho a, b, c > 0 tha :ab + cb + ca = 1. Chng minh :( )cbacba +++++++ 2111 222
Bi ton 35 : Cho a,b,c > 0 tha a+b+c =1. Chng minh :
3100111 222
++
++
+
cc
bb
aa
Bi ton 36 : Chng minh : vi mi x , y , z > 0 ta lun c :3 3 3 x y z
x y z yz zx xy
+ + + +
Bi ton 37 : Chng minh : vi a, b, c > 0 tha a + b + c = 3 ta c :
2 2 2
321 1 1
a b cb c a
+ + + + +
Bi ton 38 : Cho a, b, c 0. Chng minh :2 2 2
2 2 2 12 2 2a b c
a b b c c a+ +
+ + +
Bi ton 39 : Cho a, b, c, d dng vi a + b + c + d = 4. Hy Chng minh
a) 2 2 2 2 21 1 1 1a b c d b c d a
+ + + + + + +
b) 2 2 2 2 21 1 1 1a b c d b c c d d a a b
+ + + + + + +
c) 2 2 2 21 1 1 1 4
1 1 1 1a b c d
b c d a+ + + ++ + +
+ + + +
Bi ton 40 : Cho a, b, c > 0 v a + b + c = 3. Hy chng minh :a b c ab cb ca+ + + +
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