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A: §Æt vÊn ®Ò
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k%QL)C[^\L^C^s^]1
Bµi 1.2M
2%!K7<M
2QM
_
Z%
_
Z
_
Z!
_
ZK
_



l%ZZ!ZKm
Gi¶i :
tb+MH^
_
Z%
_
Z
_
Z!
_
ZK
_
0l%ZZ!ZKm
^l
b
a

2
m
_
Zl
c
a

2
m
_
Zl

d
a

2
m
_
Zl
e
a

2
m
_
kl
b
a

2
m
_


d@%
kl
c
a

2
m
_



d@
kl
d
a

2
m
_


d@!
kl
e
a

2
m
_

d@K
^\H

d@%!K
kqq^qqL)C[^\%^^!^K^
2
a
Bµi 1.3 :2% &M


2
22
22






+

+ baba
]d
Giải :
tb+MH^
2
22
22






+

+ baba
^
4
)2()(2

2222
bababa +++
^
0)(
4
1
)222(
4
1
22222
=+ baabbaba
1B@%1
kqq^qqL)C^%1
2. Phơng pháp 2 ; Dùng phép biến đổi tơng đơng .
0XEM=E O% &D. .@%
& $4% & P * $1
0:7<% &G!5M
lnZ=m
_
^n
_
Z_n=Z=
_
ln0=m
_
^n
_
0_n=Z=
_
lnZ=Z2m

_
^n
_
Z=
_
Z2
_
Z_n=Z_n2Z_=2
lnZ=m
c
^n
c
Zcn
_
=Zcn=
_
Z=
c
ln0=m
c
^n
c
0cn
_
=Zcn=
_
0=
c
1
B-!9M

Bài 2. 1M2%7<!.O%Q]12QM

3
4
1
1
1
1

+
+
+ ba
Giải:
k5#b#%E O. .i
clZ]Z%Z]m

glZ]ml%Z]m
r

gl%ZZ%Z]mlZ%^]m
r

g%Zf]

g%lZ%m
_


g%
= &< $1IC #)1

Bài 2. 2M2%7<!.)PMZ%Z^g
2QMlZ%ml%ZmlZm


c
%
c

c

Giải:
uMlZ%m
_


g%lZ%Zm
_
^
[ ]
cbacba )(4)(
2
+++
^\]R

glZ%m^\]RlZ%m

glZ%m
_



]R%
^\Z%

%
]]
.M%Z

%
Z

%
^\lZ%ml%ZmlZm


c
%
c

c

Bµi 2.3M2% &M

3
33
22







+

+ baba
i \di%\d
Gi¶i :
k5#b#%E O. .MB@\di%\d^\Z%\d

3
33
22






+

+ baba








+
≥+−







+
2
).(
2
22
ba
baba
ba
1
2
2






+ ba

_
0%Z%
_




2
2






+ ba
g
_
0g%Zg%
_



_
Z_%Z%
_

c
_
0R%Zc%
_


cl
_
0_%Z%

_
m

d
= &<5 $i7CM
3
33
22






+

+ baba
Bµi 2.4:
2_7<%)PZ%^]12:v
c
Z%
c
Z%


2
1
Gi¶i :
M
c

Z%
c
Z%


2
1
[^\
c
Z%
c
Z%0
2
1


d
[^\lZ%ml
_
0%Z%
_
mZ%0
2
1


d
[^\
_
Z%

_
0
2
1

d1BZ%^]
[^\_
_
Z_%
_
0]

d
[^\_
_
Z_l]0m
_
0]

dl%^0]m
[^\g
_
0gZ]

d
[^\l_0]m
_


d

= &<5 $1B>C
c
Z%
c
Z%


2
1
kqq^qqL)C^%^
2
1
]_
Bài 2.5 :2% &M
3
33
22






+

+ baba

M\d%\d1
Giải :
B@\d%\d^\Z%\d

M
3
33
22






+

+ baba
[^\
( )
2
22
22
.
2






+







+
+






+ baba
baba
ba
[^\
2
22
2






+
+
ba
baba
[^\g
_

0g%Zg%
_



_
Z_%Z%
_

[^\cl
_
0_%Z%
_
m

d
[^\cl0%m
_


d1= &C $
^\
3
33
22







+

+ baba
kqq^qqL)C^%1
Bài 2.6MB@\d%\d12% &M

a
b
a




a
b
b
Giải :
k5#b#%E O. .M

a
b
a




a
b
b

l
)() baabbbaa ++

d

[ ]
0)()()(
33
++ baabba

0)())(( +++ baabbababa

0)2)(( ++ bababa

0))(( + baba
= &< $i7CM
a
b
a




a
b
b

3. Phơng pháp 3: dùng bất đẳng thức quen thuộc .
]c
0  XE    M  k5    %  &    K     M  27  

=#L% &!?C+ < (%E O

:7<+)u% &ML
_
ZC
_


_LC
B@%\d
2≥+
a
b
b
a
2-!9M
Bµi 3.1Mw)78%7<!.QM

2>
+
+
+
+
+ ba
c
ac
b
cb
a


Gi¶i
#!9=a2CM
Zl%Zm
)(2 cba +≥

cba
a
cb
a
++

+
2
. *M

cba
b
ac
b
++

+
2

cba
c
ba
c
++


+
2
k%Q/%=a( oGL)C M
^%Z%^Z^Z%Z%Z^dl@)E% 
7<!.m1
u 7CM
2>
+
+
+
+
+ ba
c
ac
b
cb
a
Bµi 3.2:
2LC_7<)PM

L
_
ZC
_
^
22
11 xyyx −+−
 2QMcLZgC

e

Gi¶i :
¸#!9% &=#LM
lL
_
ZC
_
m
_
^l
22
11 xyyx −+−
m
_
l
1≤x
i
1≤y
m


lL
_
ZC
_
ml]0C
_
Z]0L
_
m
^\L

_
ZC
_


]
"MlcLZgCm
_


lc
_
Zg
_
mlL
_
ZC
_
m

_e
^\cLZgC

e
]g
a&L)C








=
>>
=+
43
0,0
1
22
yx
yx
yx






=
=
5
4
5
3
y
x
a+M
2
5

2
3
≤≤ x
Bµi 3. 3:2%

diZ%Z^]12QM

6≤+++++ accbba
%
5,3111 <+++++ cba
Gi¶i
¸#!9%!&=#L@_%c7<M
( )
( )
( ) ( ) ( )






+++++++≤+++++
222
1111.1.1. accbbaaccbba
^\
( )
6)22.(3
2
=++≤+++++ acbaaccbba
^\

6≤+++++ accbba
1
kqq^qqL)CM^%^^
3
1
%¸#!9% &27M

1
22
1)1(
1 +=
++
≤+
aa
a
.M
1
2
1 +≤+
b
b
i
1
2
1 +≤+
c
c
2uE/c% & *M

5,33

2
111 =+
++
≤+++++
cba
cba
k &L)C^%^^d@)EMZ%Z^ ]
B>CM
5,3111 <+++++ cba
Bµi 3.4M27<!.%)PMZ%Z^]1
2QM
9
111
≥++
cba
Gi¶i :
M
0>+
a
b
b
a
%\d
M
=++
cba
111
)
111
(

cba
++
1]^
)
111
(
cba
++
1lZ%Zm
^
111 ++++++++
b
c
a
c
c
b
a
b
c
a
b
a
]e
^
++++++ )()()(3
c
a
a
c

b
c
c
b
a
b
b
a
cZ_Z_Z_^r
^\
9
111
++
cba
kqq^qqL)CM^%^^
3
1
Bài 3.5
2LC\d12QM
yxyx +
+
411


Giải
á#!9% &27M
xyyx 2+


yx

11
+



xy
2
^\lLZCml
yx
11
+
m

g
^\
yx
11
+



yx +
4
4. Phơng pháp 4 ; Dùng các tính chất của bất đẳng thức :
0XEMk5- P * (>!9)
%>#1
2-!9M
Bài 4.1 :2_7<LC)P +MLZC^_1
2QML
g

ZC
g


_
Giải
K-%,DMlL
_
0C
_
m

dL
g
ZC
g


_L
_
C
_
_lL
g
ZC
g
m

lL
_

ZC
_
m
_
l]m
MlL0Cm
_


dL
_
ZC
_


_LC
_lL
_
ZC
_
m

lLZCm
_
_lL
_
ZC
_
m


gBMLZC^_
L
_
ZC
_


_l_m
ul]ml_mML
g
ZC
g


_
kqq^qqL)CL^C^]1
Bài 4.2:
]R
2d[%![]12QM
l]0ml]0%ml]0ml]0!m\]00%00!1
Giải :
Ml]0ml]0%m^]00%Z%
k%\d%\d^\l]0ml]0%m\]00%1
k[]]0\d^\l]0ml]0%ml]0m\l]00%ml]0m
l]0ml]0%ml]0m\]00%0ZZ%1
k%!\d]0!\diZ%\di!Z%!Z!\d
^\l]0ml]0%ml]0m\]00%0
^\l]0ml]0%ml]0ml]0!m\l]00%0ml]0!m
^\l]0ml]0%ml]0ml]0!m\]00%00!Z!Z%!Z!
^\l]0ml]0%ml]0ml]0!m\]00%00!1

Bài 4.3 :2d[%[]12QM
_
c
Z_%
c
Z_
c
[cZ
_
%Z%
_
Z
_

Giải :
k%[]^\
c
[
_
[[]i%
c
[%
_
[%[]iM
l]0
_
ml]0%m\d^\]Z
_
%\
_

Z%
^\]Z
_
%\
c
Z%
c
C
c
Z%
c
[]Z
_
%1
.M%
c
Z
c
[]Z%
_
i
c
Z
c
[]Z
_
1
^\_
c
Z_%

c
Z_
c
[cZ
_
%Z%
_
Z
_

5.phơng pháp 5 : Dùng bất đẳng thức tổng quát chứa luỹ thừa các số tự
nhiên
Bài 5.1: 2\%\d2:vM

1996 1996
1996 1996
a b
a b

+
\
1995 1995
1995 1995
a b
a b

+
w)M
a(% &% &
7E\%\d7<\

m m n n
m m n n
a b a b
a b a b

>
+ +
l]m
>>C!5#b#%E O. . (
l]m

2 2
m m m n n n
m m n n
a b b a b b
a b a b
+ +
>
+ +


]0
2 2 2 2
1
m n m n
m m n n m m n n
b b b b
a b a b a b a b
> >
+ + + +

]W
m n
m n
m n
m m n n
m m n n
m m n n
b b
b b
b b
a b a b
a b a b
b b b b
< <
+ +
+ +
1 1
1 1
m n
m n
a a
b b
<
+ +
1 1
m n
m n
a a
b b
+ > +

( ) ( )
m n
m n
m n
a a a a
b b b b
> >
l_m
= &l_m $\%\d
1
a
b
>
\>C% &l]m
$
á#!9% &
m m n n
m m n n
a b a b
a b a b

>
+ +
<\%\d\
^]rrR ^]rre % & #)x $
1996 1996
1996 1996
a b
a b


+
\
1995 1995
1995 1995
a b
a b

+
6. phơng pháp 6: Dùng bất đẳng thức về 3 cạnh của tam giác
% !%"/

[%Zl]m
%[Zl_m
[Z%lcm
uc% &O%"/7C *c% &
+"
[%Zl]m
a b c <
lgm
%[Zl_m
b c a <
lem
[Z%lcm
c a b <
lRm
Bài 6.1M
2n=2_#^Z%Zl% !"/
m12QM

2

111


+

+
cpbpap
)
111
(
cba
++
w)M
M#0^
0
2
>
+ acb
.M#0%\di#0\di
#!9E)%>#l3.5) *i
cbpapbpap
4
)()(
411
=
+


+


.M
acpbp
411


+

]f

bcpap
411


+

^\
)
111
(4)
111
(2
cbacpcpap
++≥

+

+

^\ #)1
kqq^qqL)CM#0^#0%^#0^%^1

X n=2 1
Bµi 6.2M
2% !%"/2:vM
lZ%0ml%Z0mlZ0%m

%
Gi¶i:
= &%"/E

2 2 2
0 ( )b c a a b c a− < ⇒ < − − ≤

2 2 2
0 ( )c a b b c a b− < ⇒ < − − ≤

2 2 2
0 ( )a b c c a b c− < ⇒ < − − ≤
u 
2 2 2 2 2 2 2 2 2
( ) ( ) ( )a b c b c a c a b a b c− − − − − − ≤


lZ%0ml0%Zml%0Zml%Z0ml0Z%mlZ0%m
2 2 2
a b c≤

lZ%0m
_
l%Z0m
_

lZ0%m
_
2 2 2
a b c≤

lZ%0ml%Z0mlZ0%m

%
B%%"/
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%Z0\d
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7. Ph¬ng ph¸p 7 : Chøng minh ph¶n chøng .
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Zk5+  )
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]r
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2-!9M
Bµi 7. 1 :
2d[%![]12Qi-% &77M
_l]0%m\]

c%l]0m\_
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Gi¶i:
w)78*")%< &  $1yVui
M_1c1f1c_l]0%m%l]0ml]0!m!l]0m\_1c
 ^\
[ ][ ][ ][ ]
256
1
)1()1()1()1( >−−−− ddccbbaa
l]m
:4#!9% &27M

2
1
2
1
)1( =
−+
≤−
aa
aa
^\l]0m


4
1
.M%l]0%m



4
1
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4
1
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4
1
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
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1
)1()1()1()1( >−−−− ddccbbaa
l_m
ul]ml_m7CT1
aT A-g% & D%
71
Bµi 7.2 :
lY/ ?o7C *m
2Qc7<!.%)P)%% &
7M
2
1
<+

b
a
i
2
1
<+
c
b
i
2
1
<+
a
c
_d
Gi¶i
w)78o"c7<!.%)P)c% &M

2
1
<+
b
a
i
2
1
<+
c
b
i

2
1
<+
a
c
2KuE/c% & *M

6
111
<+++++
a
c
c
b
b
a

6)
1
()
1
()
1
( <+++++
c
c
b
b
a
a

l]m
B%\dM
2)
1
( ≥+
a
a
i
2)
1
( ≥+
b
b
i
2)
1
( ≥+
c
c
^\
6)
1
()
1
()
1
( ≥+++++
c
c
b

b
a
a
aCVJ@l]m
B>Co"c7<!.%)P)c% &
1^\ #
Bµi 7.3 :
2Q7<!.%)P)c% &
7M
 gl]0%m\]ig%l]0m\]igl]0m\]1
Híng dÉn :.%_M
Bµi 7.4M
lY/ ?o7C@  $m
 2
c
Z%
c
^_12QMZ%

_1
Gi¶i :
w)78MZ%\_^\lZ%m
c
\f
^\
c
Z%
c
Zc%lZ%m\f
^\_Zc%lZ%m\flBM

c
Z%
c
^_m
^\%lZ%m\_
^\%lZ%m\
c
Z%
c
lBM
c
Z%
c
^_m
2)E7<!.% *M
%\
_
0%Z%
_
^\d\l0%m
_
BT
B>CMZ%

_
8. Ph¬ng ph¸p 8 : §æi biÕn sè
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0XEM+#.## O%E7<Q % P
!" .)..!"'% P%E)111
2-!9M

Bµi 8. 1M
2QMyE%\dM

2
3

+
+
+
+
+ ab
c
ac
b
cb
a
Gi¶i:
a4M%Z^LZ^CZ%^s
^\Z%Z^
2
zyx ++
^\^
2
xzy −+
%^
2
yxz −+
^
2
zyx −+

X M
B^
ab
c
ac
b
cb
a
+
+
+
+
+
^
z
zyx
y
yxz
x
xzy
222
−+
+
−+
+
−+
^
2
3
2

3
111
2
3
)(
2
1
)(
2
1
)(
2
1
=−++≥−+++++
z
y
y
z
z
x
x
z
y
x
x
y
Bµi 8.2M
2Qi@7<LC% &M
 0
4

1
)1()1(
)1)((
4
1
2222
2222

++


yx
yxyx
Gi¶iM
a4M^
)1)(1(
22
22
yx
yx
++

%^
)1)(1(
1
22
22
yx
yx
++


^\%^
2222
2222
)1()1(
)1)((
yx
yxyx
++
−−
!zC@%M0
22
)(
4
1
)(
4
1
baabba +≤≤−
 :Ml0%m
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^
2
2
1
2
1







+

x
lZ%m
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^
2
2
1
2
1






+

y
ICM0
4
1


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4
1
1
Bµi 8.3M
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2
1
2
1
2
1
222

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111
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9
111
++
zyx
1
9.Phơng pháp 9: Dùng phép quy nạp toán học .
0XEMa(% & $@\]%Q#0
.##C"#EM
ZX(% & $@^]l^
d
m
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ZXE>% & $@\]l\
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2Q@7<C!.

c
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ZB@^cM_


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l{m $@^c1
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1
4
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1
6
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111
n
n
2
12


13
1
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l{ml7<C!.m
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2
1
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4
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111
k
k
2
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1
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Dl{m $@^Z]M

2
1

1
4
3
1
6
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111
k
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2
12
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13
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12
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10.Phơng pháp 10 : Chứng minh bất đẳng thức trong hình học phẳng
Bài 10.1M2:vOCE/@
.gD%- G"E#
G
C1
B
A
C
0
A1
B1
Giải:
w% !% GCEv%- G
"E#

n=2#)Z%Z\gv
_g

B

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n=2EwVn=2VdQF
%wn=wn2w=21w)78Vd
Qwn=dnZd=^_vwnZw=\_vwn^
2
3
nn
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2
3
w=^
2
3
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2
3
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ywnZw=\_v

2
3
lZ%m\_v

Z%\cv
:d22

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\v
k Z%Z\cvZv^gv1
B>CZ%Z\gv
Bµi 10. 2M: GE#L$@"/ |
n" (=2jE#CE@ G,"n=n2
":yQ
3
AB AC+
<
:=Zy2[
2
AB AC+
Gi¶i

B
C
l
0
A
M
N
wxE# (/E#CE:y@ GV
d-E#C
:=^:xy2^yx
u :y^:=Zy2n:y:y[n:Zny

y_:y[n:ZnyZ=:Z2y^n=Zn2

:y[
2
AB AC+
yn:y;"C:y\n:
:y\ny

_:y\n:Zny
B:y^=2Z2y
yc:y\n:ZnyZ=:Z2y! c:y\n=Zn2

:y\
3
AB AC+
B>C
3
AB AC+
<
:=Zy2[
2
AB AC+
_e
11 . Ngoài ra còn có một số phơng pháp khác để chứng minh bất đẳng
thức nh : Phơng pháp làm trội , tam thức bậc hai ta phải căn cứ vào
đặc thù của mỗi bài toán mà sử dụng phơng pháp cho phù hợp . Trong
phạm vi nhỏ của đề tài này không hệ thống ra những phơng pháp đó .
iii : ứng dụng của bất đẳng thức
1- Dùng bất đẳng thức để tìm cực trị .
0XEMyEhlLm


hlLm?A1
yEhlLm

:hlLm?@:1
G C # !9 % & !9 M 27
=#L% &!?C+ <1
X(G*#L)C! & (?1
?/%(!" C78!9#.
##%E O. . O%E7<7<% &111
?/%(!?C+ <>!9
% &!?C+ <
2$TM
BABA ++

t)C!qq^qqn=

d

0A
kqq^qqL)Cn^d
Bài 1 :?A/%(M=^
c
Z%
c
Z%i2%E%
)PMZ%^]1
Giải
=^lZ%ml
_

0%Z%
_
mZ%
^
_
0%Z%
_
Z%^
_
Z%
_

M_l
_
Z%
_
m

lZ%m
_
^]^\
_
Z%
_



2
1
B>C=^

2
1
^%^
2
1
Bài 2M?A/%(M
n^lL
_
ZLmlL
_
ZL0gm
%?A/%(M
=^0L
_
0C
_
ZLCZ_LZ_C
Giải
n^lL
_
ZLmlL
_
ZL0gm1a4M^L
_
ZL0_
^\n^l0_mlZ_m^
_
0g

0g

k%QL)CM^dL
_
ZL0_^d
_R
lL0_mlLZ_m^dL^0_iL^]1
^\n^0gL^0_iL^]i
%.

Bµi 3 : ?A/%(1
2^
1232 −+− xx

%k^
63
22
−++++ xxxx

}^
4321 −+−+−+− xxxx
Gi¶i :
¸#!9=aM
BABA +≥+
kqq^qqL)Cn=

d1
^\2^
2221322132 =−=−+−≥−+− xxxx
kqq^qqL)Cl_L0cml]0_Lm

d

2
3
2
1
≤≤ x

B>C2^_
2
3
2
1
≤≤ x
%.Mk^rM0c

L

_
}^gM_

L

c
Bµi 4 :2[%[[!M
:hlLm^
ax −
Z
bx −
Z
cx −
Z

dx −
Híng dÉnM.MhlLm^!Z0%0%

L


Bµi 5M2%7<!.LCs)PM
x+1
1
Z
y+1
1
Z
z+1
1


_
?@/-MY^LCs
Gi¶iM

x+1
1


l]0
y+1
1
mZl]0
z+1

1
m^
y
y
+1
Z
z
z
+1


_
)1)(1( zy
yz
++
.M
y+1
1


_
)1)(1( zx
zx
++

z+1
1


_

)1)(1( yx
xy
++
u 7CMY^LCs


8
1
:LY^
8
1
L^C^s^
2
1
_W
Bµi 6 : 2c7<!.%)PMZ%Z^]1?A
/%(M~^
222
)
1
()
1
()
1
(
c
c
b
b
a

a +++++

Gi¶i:
M~^l
_
Z%
_
Z
_
mZl
222
111
cba
++
mZR
B>!9% &=#LM
l1]Z%1]Z1_m
_


cl
_
Z%
_
Z
_
m
^\
_
Z%

_
Z
_



3
1
.M
2
)
111
(
cba
++


c
)
111
(
222
cba
++
:4M
=++
cba
111
l
cba

111
++
m1]^l
cba
111
++
mlZ%Zm
^cZl
a
b
b
a
+
mZl
b
c
c
b
+
mZl
c
a
a
c
+
m

cZ_Z_Z_^r
^\
cba

111
++


r
^\
2
)
111
(
cba
++


f]
^\
)
111
(
222
cba
++


_W
~


3
1

Z_WZR^cc
Dấu '' = '' xảy ra khi : a = b = c =
3
1
Vậy MinF = cc
3
1
M^%^^
3
1
.
Bài 7 : Cho G =
xyz
zxyyzxxyz 321 −+−+−
Tìm giá trị lớn nhất của G :
Giải : Tập xác định : x

]iC

_is

c
M G =
x
x 1−
+
y
y 2−
+
z

z 3−
Theo BĐT Côsi ta có :
2
11
1
+−
≤−
x
x
=>
x
x 1−

2
1

_f
.M
22
1
2


y
y
i
32
13



z
z
^\w


32
1
22
1
2
1
++
B>C:Lw^
32
1
22
1
2
1
++
" *L^_iC^_is^R
Bài 8?A/H^
1x
x
@L\]1
%1?@/X^
2
1. xx

HDM#!9% &27.%eM

2 - Dùng bất đẳng thức để giải phơng trình .
0XEMyG-/% &#.##
% &%E OElBBYm/#.7
7C> (|+/#.1
yEB^BY"47<? /l)P
tam
^\#.+1
yEB\BY4B[BY"?/1
^\#.+1
02-!9M
Bài 1Mw)#.M
]c
1x
Zr
1+x
^]RL
GiảiM
a+ML

]l{m
2]M#!9% &27M]c
1x
Zr
1+x
^]c1_1
1
2
1
x
Zc1_1

1
2
3
+x


]clL0]Z
4
1
mZclLZ]Z
4
9
m^]RL
kqq^qqL)C






=+
=
2
3
1
2
1
1
x
x

L^
4
5
)Pl{m
_r
Y.l]m+!qq^qqFl_mL)C
B>Cl]m+L^
4
5
1

Bài 2M?@/p^
32 x
Z
x25
%1w)#.M
32 x
Z
x25
0L
_
ZgL0R^dl{m
Giải :
1,Ml
32 x
Z
x25
m
_



_l_L0cZe0_Lm^g

32 x
Z
x25


_
^\:Lp^_L^_1
%1taM
2
5
2
3
x

l{m
32 x
Z
x25
^L
_
0gLZR
BY^lL0_m
_
Z_

_!qq^qqL)CL^_1
^\@L^_l)PtamB^BY^_1

^\#.l{m+L^_1
Bài 3 :w)#.M

x6
Z
2+x
^L
_
0RLZ]c
Giải : taM0_

L

R1
BY^lL0cm
_
Zg

g1kqq^qqL)CL^c1
B
_
^l
x6
1]Z
2+x
1]m
_


lR0LZLZ_ml]Z]m^]R

^\B

g!qq^qqL)C
x6
^
2+x
L^_1
^\?/L (B^BY^\Y.+
Bài 4Mw)#.M

16123
2
+ xx
Z
134
2
+ yy
^e
HD M
16123
2
+ xx

_i
134
2
+ yy

c^\B


e1
kqq^qqL)CM



=
=
02
02
y
x




=
=
2
2
y
x

^\#.+ML^_iC^_1

3 - Dùng bất đẳng thức để giải hệ phơng trình :
cd

×