G.NTH
1. C¸c kiÕn thøc cÇn n¾m
1.1. C¸c hÖ thøc c¬ b¶n
+ cos 2 α + sin 2 α = 1
1
π
(α ≠ + kπ)
2
2
cos α
1
+ 1 + cotg2α =
(α ≠ kπ)
sin 2 α
+ 1 + tg2α =
kπ
)
2
1.2. C«ng thøc céng gãc
+ cos(α ± β) = cosα cosβ sinα sinβ
+ sin(α ± β) = sinα cosβ ± cosα sinβ
tgα ± tgβ
π
+ tg (α ± β) =
(α ; β ≠ + kπ)
1 tgα tgβ
2
cot gα. cot gβ 1
+ cotg(α ± β) =
(α; β ≠ kπ)
cot gα ± cot gβ
1.3. C«ng thøc nh©n
+ sin2α = 2 sinα cosα
+ cos2α = cos2α - sin2α = 2cos2α - 1 = 1 - 2sin2α
2 tgα
π
π
+ tg2α =
(α ≠ + k )
2
4
2
1 − tg α
+ tgα . cotgα = 1 (α ≠
cot g 2 α − 1
kπ
(α ≠ )
2 cot gα
2
+ sin3α = 3sinα - 4sin3α
+ cos3α = 4cos3α - 3cosα
3tgα − tg 3α
π
π
+ tg3α =
(α ≠ + k )
3
6
3
1 − 3tg α
1.4. C«ng thøc h¹ bËc
1 + cos 2α
1 − cos 2α
+ cos2α =
+ sin2α =
2
2
π
1 − cos 2α
(α ≠ + kπ)
+ tg2α =
2
1 + cos 2α
1.5. C«ng thøc biÕn ®æi tæng thµnh tÝch:
α+β
α −β
cos
+ cosα + cosβ = 2cos
2
2
α +β
α β
sin
+ cosα - cosβ = - 2sin
2
2
α+β
α β
cos
+ sinα + sinβ = 2sin
2
2
α +β
α −β
sin
+ sinα - sinβ = = - 2cos
2
2
+ cotg2α =
1
G.NTH
sin(α ± β)
π
(α; β ≠ + kπ)
cos α. cos β
2
1.6. C«ng thøc biÕn ®æi tÝch thµnh tæng:
1
+ cosα.cosβ = [cos(α + β) + cos(α − β)]
2
1
+ sinα.sinβ = [cos(α − β) + cos(α + β)]
2
1
+ sinα.cosβ = [sin(α + β) + sin(α − β)]
2
+ tgα ± tgβ =
BiÓu thøc ®¹i sè
BiÓu thøc lîng gi¸c
t¬ng tù
1 + x2
1 + tan2t
4x3 - 3x
2x2 - 1
2x
1− x2
2x
1+ x2
x+y
1 − xy
4cos3t - 3cost
2cos2t - 1
1
cos 2 t
4cos3t - 3cost = cos3t
2cos2t - 1 = cos2t
2 tan t
1 − tan 2 t
2 tan t
= tan2t
1 − tan 2 t
2 tan t
1 + tan 2 t
2 tan t
= sin2t
1 + tan 2 t
tan + tan
1 − tan tan
tan + tan
= tan(α+β)
1 − tan tan
C«ng thøc lîng gi¸c
1+tan2t =
1
1
−1
− 1 = tan2α
2
2
cos α
cos α
...
....
......
mét sè ph¬ng ph¸p lîng gi¸c ®Ó chøng minh
bÊt ®¼ng thøc ®¹i sè
x2 - 1
I. D¹ng 1: Sö dông hÖ thøc sin2 + cos2 = 1
1) Ph¬ng ph¸p:
x = sin α
a) NÕu thÊy x2 + y2 = 1 th× ®Æt
víi α ∈ [0, 2π]
y = cos α
x = r sin
víi α ∈ [0, 2π]
y = r cos
b) NÕu thÊy x2 + y2 = r2 (r > 0) th× ®Æt
2. C¸c vÝ dô minh ho¹:
VD1: Cho 4 sè a, b, c, d tho¶ m·n: a2 + b2 = c2 + d2 = 1
Chøng minh r»ng: − 2 ≤ a(c+d) + b(c-d) ≤ 2
2
G.NTH
Gi¶i:
c = sin v
a = sin u
vµ
⇒ S = sinu(sinv+cosv) + cosu(sinv-cosv)
b = cos u
d = cos v
§Æt
⇒ P = a(c+d) + b(c-d) = (sinucosv+cosusinv) - (cosucosv - sinusinv)
= sin(u+v) - cos(u+v)
π
⇔ S = 2 sin(u + v) − ∈[− 2, 2] ⇒ − 2 ≤ S = a(c + d) + b(c − d) ≤ 2 (®pcm)
4
2
2
1
1
25
VD2: Cho a + b = 1. Chøng minh r»ng: a 2 + 2 + b 2 + 2 ≥
2
a
b
2
2
Gi¶i:
§Æt a = cosα vµ b = sinα víi 0 ≤ α ≤ 2π. ThÕ vµo biÓu thøc vÕ tr¸i råi biÕn ®æi.
2
2
2
1 2
1
2 1 2 1 2
a + 2 + b + 2 = cos α +
+ sin α + 2
2
a
b
cos α
sin α
2
1
1
cos 4 α + sin 4 α
4
4
+
+
4
=
cos
α
+
sin
α
+
+4
cos 4 α sin 4 α
cos 4 α. sin 4 α
= cos4α + sin4α +
(
)
[(
)
1
= cos 4 α + sin 4 α 1 +
+4
4
4
cos α. sin α
]
1
= cos 2 α + sin 2 α − 2 cos 2 α sin 2 α 1 +
+4
4
4
cos α. sin α
16
17
25
1
1
= 1 − sin 2 2α 1 + 4
(®pcm)
+ 4 ≥ 1 − (1 + 16) + 4 = + 4 =
2
2
2
sin 2α
2
B©y giê ta ®Èy bµi to¸n lªn møc ®é cao h¬n mét bíc n÷a ®Ó xuÊt hiÖn a2+b2=1
VD3: Cho a2 + b2 - 2a - 4b + 4 = 0. Chøng minh r»ng:
A = a 2 − b 2 + 2 3ab − 2(1 + 2 3 )a + (4 − 2 3 )b + 4 3 − 3 ≤ 2
Gi¶i:
BiÕn ®æi ®iÒu kiÖn: a2 + b2 - 2a - 4b + 4 = 0⇔ (a-1)2 + (b-2)2 = 1
a − 1 = sin α
a = 1 + sin α
§Æt
⇒
⇒ A = sin 2 α − cos 2 α + 2 3 sin α cos α
b − 2 = cos α b = 2 + cos α
A = 3 sin 2α − cos 2α = 2
3
1
π
sin 2α − cos 2α = 2 sin( 2α − ) ≤ 2 (®pcm)
2
2
6
VD4: Cho a, b tho¶ m·n : 5a + 12b + 7 = 13
3
G.NTH
Chøng minh r»ng: a2 + b2 + 2(b-a) ≥ - 1
Gi¶i:
BiÕn ®æi bÊt ®¼ng thøc: a2 + b2 + 2(b-a) ≥ - 1 ⇔ (a-1)2 + (b + 1)2 ≥ 1
a − 1 = R sin α
§Æt
víi R ≥ 0 ⇔
b + 1 = R cos α
a = R sin α + 1
⇔ (a − 1) 2 + (b + 1) 2 = R 2
b = R cos α − 1
Ta cã: 5a + 12b + 7 = 13 ⇔ 5(R sin α + 1) + 12(R cos α − 1) + 7 = 13
⇔ 5R sin α + 12R cosα = 13 ⇔ 1 = R
5
12
5
sin α + cosα = R sin α + arccos ≤ R
13
13
13
Tõ ®ã ⇒ (a-1)2 + (b+1)2 = R2 ≥ 1 ⇔ a2 + b2 + 2(b - a) ≥ - 1 (®pcm)
II. D¹ng 2: Sö dông tËp gi¸ trÞ | sin α |≤ 1 ; | cos α | ≤ 1
1. Ph¬ng ph¸p:
x = sin khi ∈ − 2 ; 2
x = cos khi ∈ [ 0; ]
a) NÕu thÊy |x| ≤ 1 th× ®Æt
x = m sin khi ∈ − 2 ; 2
b) NÕu thÊy |x| ≤ m ( m ≥ 0 ) th× ®Æt
x = m cos khi ∈ [ 0; ]
2. C¸c vÝ dô minh ho¹:
VD1: Chøng minh r»ng: (1+x)p + (1-x)p ≤ 2p ∀ |x| ≤ 1 ; ∀ P ≥ 1.
Gi¶i:
§Æt x = cosα víi α ∈ [0, π], khi ®ã (1 + x)p + (1 - x)p = (1+cosα)p + (1-cosα)p
p
p
α
α
α
α
α
α
= 2 cos 2 + 2 sin 2 = 2 p cos 2 p + sin 2 p ≤ 2 p cos 2 + sin 2 = 2 p
2
2
2
2
2
2
(®pcm)
VD2: Chøng minh r»ng:
3−2
3+2
≤ 3x 2 + x 1 − x 2 ≤
2
2
Gi¶i:
Tõ ®k 1 - x2 ≥ 0 ⇔ |x| ≤ 1 nªn
§Æt x = cosα víi 0 ≤ α ≤ π ⇒ 1 − x 2 = sinα. Khi ®ã ta cã:
P= 2 3 x 2 + 2 x 1 − x 2 = 2 3 cos 2 + 2 cos sin = 3 (1 + cos 2 ) + sin 2
4
G.NTH
3
1
π
= 2 cos2α + sin 2α + 3 = 2 sin 2α + + 3 ⇒ 3 − 2 ≤ A ≤ 3 + 2 (®pcm)
2
3
2
VD3: Chøng minh r»ng: 1 + 1 − a 2
[ (1 + a)
3
]
− (1 − a )3 ≤ 2 2 + 2 − 2a 2 (1)
Gi¶i:
Tõ ®k |a| ≤ 1 nªn
§Æt a=cosα víi α∈[0,π] ⇒ 1 − a = 2 sin
(1)⇔
1 + 2 sin
α
α
; 1 + a = 2 cos ; 1 − a 2 = sin α
2
2
α
α
α
α
α
α
cos .2 2 cos 3 − sin 3 ≤ 2 2 + 2 2 sin cos
2
2
2
2
2
2
α
α
α
α
α
α
α
α
α
α
⇔ sin + cos cos − sin cos2 + sin cos + sin 2 ≤ 1 + sin cos
2
2
2
2
2
2
2
2
2
2
α
α
α
α
α
α
⇔ sin + cos cos − sin = cos 2 − sin 2 = cos α ≤ 1 ®óng ⇒ (®pcm)
2
2
2
2
2
2
) (
(
)
VD4: Chøng minh r»ng: S = 4 (1 − a 2 )3 − a 3 + 3 a − 1 − a 2 ≤ 2
Gi¶i:
Tõ ®k |a| ≤ 1 nªn:
§Æt a = cosα víi α ∈ [0, π] ⇒ 1 − a 2 = sinα. Khi ®ã biÕn ®æi S ta cã:
S= 4(sin 3 α − cos 3 α) + 3(cos α − sin α) = (3 sin α − 4 sin 3 α) + (4 cos 3 α − 3 cos α)
π
= sin 3α + cos 3α = 2 sin 3α + ≤ 2 ⇒ (®pcm)
4
(
)
VD5: Chøng minh r»ng A = a 1 − b 2 + b 1 − a 2 + 3 ab − (1 − a 2 )(1 − b 2 ) ≤ 2
Gi¶i:
Tõ ®iÒu kiÖn: 1 - a2 ≥ 0 ; 1 - b2 ≥ 0 ⇔ |a| ≤ 1 ; |b| ≤ 1 nªn.
π π
§Æt a = sinα, b = sin β víi α, β ∈ − ;
2 2
Khi ®ã A = sin α cos β + cos α sin β − 3 cos(α + β) =
1
3
π
= sin(α + β) − 3 cos(α + β) = 2 sin(α + β) − cos(α + β) = 2 sin(α + β) − ≤ 2
2
2
3
(®pcm)
VD6: Chøng minh r»ng: A = |4a3 - 24a2 + 45a - 26| ≤ 1 ∀a ∈ [1; 3]
5
G.NTH
Gi¶i:
Do a ∈ [1, 3] nªn |a-2| ≤ 1 nªn ta ®Æt a - 2 = cosα ⇔ a = 2 + cosα. Ta cã:
A = 4(2 + cosα)3 − 24(2 + cosα)2 + 45(2 + cosα) − 26 = 4cos3 α − 3cosα = cos3α ≤1
(®pcm)
VD7: Chøng minh r»ng: A =
2a − a 2 − 3a + 3 ≤ 2 ∀ a ∈[0, 2]
Gi¶i:
Do a ∈ [0, 2] nªn |a-1| ≤ 1 nªn ta ®Æt a - 1 = cosα víi α ∈ [0, π]. Ta cã:
A=
2(1 + cos α ) − (1 − cos α ) 2 − 3 (1 + cos α ) + 3 = 1 − cos 2 α − 3 cos α
1
3
π
cos α = 2 sin α + ≤ 2 (®pcm)
= sin α − 3 cos α = 2 sin α −
2
3
2
III. D¹ng 3: Sö dông c«ng thøc: 1+tg2 =
π
1
1
2
(
α
≠
+ kπ)
⇔
tg
α
=
−
1
2
cos2 α
cos2 α
1) Ph¬ng ph¸p:
a) NÕu |x| ≥ 1 hoÆc bµi to¸n cã chøa biÓu thøc
th× ®Æt x =
1
π 3π
víi α∈ 0; ∪ π,
cos α
2 2
b) NÕu |x| ≥ m hoÆc bµi to¸n cã chøa biÓu thøc
th× ®Æt x =
x2 −1
x 2 − m2
m
π 3π
víi α∈ 0; ∪ π,
cos α
2 2
2. C¸c vÝ dô minh ho¹:
VD1: Chøng minh r»ng A =
a2 −1 + 3
≤ 2 ∀ a ≥1
a
Gi¶i:
Do |a| ≥ 1 nªn :
§Æt a =
A=
1
π 3π
víi α∈ 0; ∪ π, ⇒
cos α
2 2
a 2 − 1 = tg 2 α = tgα . Khi ®ã:
a 2 −1 + 3
π
= (tgα + 3) cosα = sin α + 3 cosα = 2 sin α + ≤ 2 (®pcm)
a
3
5 − 12 a 2 − 1
VD2: Chøng minh r»ng: - 4 ≤ A =
≤ 9 ∀ a ≥1
a2
Gi¶i:
6
G.NTH
Do |a| ≥ 1 nªn:
§Æt a =
1
π 3π
víi α∈ 0; ∪ π, ⇒
cos α
2 2
a 2 − 1 = tg 2 α = tgα . Khi ®ã:
5(1+ cos2α)
5−12 a2 −1
− 6sin2α
= (5-12tgα)cos2α = 5cos2α-12sinαcosα=
2
2
a
5 13 5
12
5
5 13
= + cos 2α − sin 2α = + cos 2α + arccos
2 2 13
13
13
2 2
A =
⇒-4=
5 13
5 13
5 5 13
+ (−1) ≤ A = + cos 2α + arccos ≤ + .1 = 9 (®pcm)
2 2
2 2
13 2 2
VD3: Chøng minh r»ng: A =
a 2 − 1 + b2 − 1
≤1
ab
∀ a ; b ≥1
Gi¶i:
Do |a| ≥ 1; |b| ≥ 1 nªn .
§Æt a =
1
1
π 3π
;b=
víi α∈ 0; ∪ π, . Khi ®ã ta cã:
cos β
cos α
2 2
A = ( tgα + tgβ) cos α cos β = sin α cos β + sin β cos α = sin(α + β) ≤ 1 (®pcm)
VD4: Chøng minh r»ng: a +
a
a −1
2
≥ 2 2 ∀ a >1
Gi¶i:
Do |a| > 1 nªn:
§Æt a =
a+
1
a
1
1
1
π
víi α∈ 0; ⇒
. Khi ®ã:
=
.
=
cos α
2
a 2 − 1 cos α tg 2 α sin α
a
a2 −1
=
1
1
1
1
2 2
+
≥ 2.
.
=
≥ 2 2 (®pcm)
cos α sin α
cos α sin α
sin 2α
VD5: Chøng minh r»ng y x 2 − 1 + 4 y 2 − 1 + 3 ≤ xy 26 ∀ x ; y ≥ 1
Gi¶i:
BÊt ®¼ng thøc ⇔
x2 − 1
+
x
Do |x|; |y| ≥ 1 nªn §Æt x =
1 4 y 2 − 1 3
+
≤ 26 (1)
x
y
y
1
1
π
; y=
víi α, β∈ 0, .
cosβ
cos α
2
7
G.NTH
Khi ®ã: (1) ⇔ S = sinα + cosα(4sinβ + 3cosβ) ≤
26
Ta cã: S ≤ sinα + cosα (4 2 + 32 )(sin 2 β + cos 2 β) = sin α + 5 cos α
≤ (12 + 52 )(sin 2 + cos 2 ) = 26 ⇒ (®pcm)
IV. D¹ng 4: Sö dông c«ng thøc 1+ tg2 =
1
cos 2 α
1. Ph¬ng ph¸p:
π π
a) NÕu x ∈ R vµ bµi to¸n chøa (1+x2) th× ®Æt x = tgα víi α ∈ − ,
2 2
π π
b) NÕu x ∈ R vµ bµi to¸n chøa (x2+m2) th× ®Æt x = mtgα víi α ∈ − ,
2 2
2. C¸c vÝ dô minh ho¹:
VD1: Chøng minh r»ng: S =
3x
1 + x2
−
4x 3
(1 + x 2 )3
≤1
Gi¶i:
1
π π
§Æt x = tgα víi α ∈ − , ⇒ 1 + x 2 =
, khi ®ã biÕn ®æi S ta cã:
cos α
2 2
S = |3tgα.cosα - 4tg3α.cos3α| = |3sinα - 4sin3α| = |sin3α| ≤ 1
(®pcm)
3 + 8a 2 + 12a 4
VD2: T×m gi¸ trÞ lín nhÊt vµ nhá nhÊt cña biÓu thøc A =
(1 + 2a 2 ) 2
Gi¶i:
3 + 4 tg 2 α + 3tg 4 α
π π
§Æt a 2 = tgα víi α∈ − , th× ta cã: A =
(1 + tg 2 α) 2
2 2
3 cos 4 α + 4 sin 2 α cos 2 α + 3 sin 4 α
=
= 3(sin 2 α + cos 2 α) 2 − 2 sin 2 α cos 2 α
2
2
2
(cos α + sin α)
sin 2 2α
5
1
sin 2 2α
0
⇒ = 3− ≤ A = 3−
≤ 2− =3
2
2
2
2
2
1
π
5
Víi α = 0 ⇒ a = 0 th× MaxA = 3 ; Víi α = ⇒ a =
th× MinA =
2
4
2
=3-
VD3: Chøng minh r»ng:
(a + b)(1 − ab) 1
≤ ∀ a, b ∈ R
(1 + a 2 )(1 + b 2 ) 2
Gi¶i:
8
G.NTH
§Æt a = tgα, b = tgβ. Khi ®ã
= cos 2 α cos 2 β.
(a + b )(1 − ab)
(tgα + tgβ)(1 − tgαtgβ)
=
2
2
(1 + a )(1 + b )
(1 + tg 2 α)(1 + tg 2β)
sin(α + β) cos α. cos β − sin α. sin β
.
cos α. cos β
cos α. cos β
1
1
sin[2(α + β)] ≤ (®pcm)
2
2
| a −b |
| b−c|
| c −a |
VD4: Chøng minh r»ng:
+
≥
∀a, b,c
(1+a2)(1+b2) (1+b2)(1+c2)
(1+c2)(1+a2)
= sin(α + β) cos(α + β) =
Gi¶i:
§Æt a = tgα, b = tgβ, c = tgγ. Khi ®ã bÊt ®¼ng thøc ⇔
| tg α − tg β |
| tg β − tg γ |
| tg γ − tg α |
⇔
+
≥
(1 + tg 2 α )(1 + tg 2 β )
(1 + tg 2 β )(1 + tg 2 γ )
(1 + tg 2 γ )(1 + tg 2 α )
⇔ cos α cos β.
sin(α − β)
sin(β − γ )
sin( γ − α)
+ cos β cos γ.
≥ cos γ cos α.
cos α. cos β
cos β. cos γ
cos γ. cos α
⇔ |sin(α-β)|+|sin(β-γ)| ≥ |sin(γ-α)|. BiÕn ®æi biÓu thøc vÕ ph¶i ta cã:
|sin(γ-α)|= |sin[(α-β)+(β-γ)]| = |sin(α-β)cos(β-γ)+sin(β-γ)cos(α-β)| ≤
|sin(α-β)cos(β-γ)|+|sin(β-γ)cos(α-β)|=|sin(α-β)||cos(β-γ)|+|sin(β-γ)||cos(α-β)|
≤ |sin(α-β)|.1 + |sin(β-γ)|.1 = |sin(α-β)| + |sin(β-γ)| ⇒ (®pcm)
ab + cd ≤ (a + c)(b + d ) (1) ∀a , b, c, d > 0
VD5: Chøng minh r»ng:
Gi¶i:
(1) ⇔
ab
+
( a + c )( b + d )
cd
≤1⇔
( a + c )( b + d )
cd
1
ab
+
≤1
c
b
c
b
1 + 1 +
1 + 1 +
a d
a d
c
d
π
§Æt tg2α= , tg2β= víi α,β ∈ 0, ⇒ BiÕn ®æi bÊt ®¼ng thøc
a
b
2
⇔
1
(1 + tg α)(1 + tg β)
2
2
+
tg2α.tg2β
(1 + tg α)(1 + tg β)
2
2
= cos2 α cos2 β + sin2 α sin2 β ≤ 1
⇔ cosα cosβ + sinα sinβ = cos(α-β) ≤ 1 ®óng ⇒ (®pcm)
DÊu b»ng x¶y ra ⇔ cos(α-β) = 1 ⇔ α=β ⇔
c d
=
a b
6a + 4 | a 2 − 1 |
VD6: T×m gi¸ trÞ lín nhÊt vµ nhá nhÊt cña biÓu thøc A =
a2 +1
9
G.NTH
Gi¶i:
§Æt a = tg
α
. Khi ®ã A =
2
6 tg
α
α
α
α
+ 4 | tg 2 − 1 |
2 tg
tg 2 − 1
2
2 + 4.
2
2
= 3.
α
α
α
tg 2 + 1
1 + tg 2
tg 2 + 1
2
2
2
A = 3sin α + 4 |cosα| ≥ 3 sinα + 4.0 = 3sinα ≥ 3.(-1) = -3
Sö dông bÊt ®¼ng thøc Cauchy - Schwarz ta cã:
A2 = (3sinα + 4 |cosα|)2 ≤ (32 + 42)(sin2α + cos2α) = 25 ⇒ A ≤ 5
Víi sinα = 1 ⇔ a = 1 th× MinA = - 3 ; víi
sin α | cos α |
th× MaxA = 5
=
3
4
V. D¹ng 5: §æi biÕn sè ®a vÒ bÊt ®¼ng thøc tam gi¸c
1) Ph¬ng ph¸p:
π
x; y; z > 0
A; B; C ∈ (0; )
a) NÕu 2
th× ∃∆ABC :
2
2
2
x
+
y
+
z
+
2
xyz
=
1
x = cos A; y = cos B; z = cos C
π
x; y; z > 0
A; B; C ∈ (0; )
b) NÕu
th× ∃∆ABC :
2
x + y + z = xyz
x = tgA; y = tgB; z = tgC
π
A; B; C ∈ (0; 2 )
x; y, z > 0
x = cot gA; y = cot gB; z = cot gC
c) NÕu
th× ∃∆ABC :
A; B; C ∈ (0; π)
xy + yz + zx = 1
A
B
C
x
=
tg
;
y
=
tg
;
z
=
tg
2
2
2
2. C¸c vÝ dô minh ho¹:
VD1: Cho x, y, z > 0 vµ zy + yz + zx = 1. T×m gi¸ trÞ nhá nhÊt cña biÓu thøc.
1 1 1
S = + + − 3( x + y + z)
x y z
Gi¶i:
Tõ 0 < x, y, z < 1 nªn ®Æt x = tg
Do xy + yz + zx = 1 nªn tg
α
β
γ
π
; y = tg ; z = tg víi α, β, γ ∈ 0,
2
2
2
2
α β
β γ
γ α
tg + tg tg + tg tg
=1
2 2
2 2
2 2
10
G.NTH
β
γ
tg + tg
α β
β γ
γ
2 = 1 ⇔ tg β + γ = cot g α
⇔ tg tg + tg = 1 - tg tg ⇔ 2
2
2 2
2 2 1 − tg β tg γ tg α
2
2 2
2 2
2
β γ π α
α+β+ γ π
β γ
π α
= ⇔ α+β+ γ = π
⇔ tg + = tg + ⇔ + = − ⇔
2 2 2 2
2
2
2 2
2 2
S=
1 1 1
α
β
γ α
β
γ
+ + − 3( x + y + z) = cotg + cotg + cotg -3 tg + tg + tg
2
2 2
x y z
2
2
2
α
α
β
β
γ
γ α
β
γ
S = cot g − tg + cot g − tg + cot g − tg − 2 tg + tg + tg
2
2
2
2
2
2 2
2
2
β
γ
α
S = 2(cotgα+cotgβ+cotgγ) - 2 tg + tg + tg
2
2
2
γ
α
β
S = (cotgα+cotgβ-2tg ) + (cotgβ+cotgγ-2tg ) +(cotgα+cotgβ-2tg )
2
2
2
§Ó ý r»ng: cotgα + cotgβ =
sin(α + β)
2 sin γ
2 sin γ
=
=
sin α. sin β 2 sin α. sin β cos(α − β) − cos(α + β)
γ
γ
4 sin cos
2 sin γ
2 sin γ
2
2 = 2 tg γ ⇒ cot gα + cot gβ − 2 tg γ ≥ 0
=
=
≥
γ
1 − cos(α + β) 1 + cos γ
2
2
2 cos 2
2
T ®ã suy ra S ≥ 0. Víi x = y = z =
VD2: Cho 0 < x, y, z < 1 vµ
1
th× MinS = 0
3
x
y
z
4 xyz
+
+
=
2
2
2
2
1− x 1− y 1− z
(1 − x )(1 − y 2 )(1 − z 2 )
T×m gi¸ trÞ nhá nhÊt cña biÓu thøc S = x2 + y2 + z2
Gi¶i:
Do 0 < x, y, z < 1 nªn ®Æt x = tg
Khi ®ã tgα =
⇔
α
β
γ
π
; y = tg ; z = tg víi α, β, γ ∈ 0,
2
2
2
2
2x
2y
2z
;
tgβ
=
;
tgγ
=
vµ ®¼ng thøc ë gi¶ thiÕt
1− x2
1 − y2
1 − z2
2x
2y
2z
8xyz
+
+
=
⇔ tgα+tgβ+tgγ = tgα.tgβ.tgγ
2
2
2
2
1− x 1− y 1− z
(1 − x )(1 − y 2 )(1 − z 2 )
11
G.NTH
⇔ tgα + tgβ = - tgγ(1-tgα.tgβ) ⇔
tgα + tgβ
= - tgγ ⇔ tg(α+β) = tg(-γ)
1 − tgα.tgβ
π
Do α, β, γ ∈ 0, nªn α + β = π - γ ⇔ α + β + γ = π. Khi ®ã ta cã:
2
tg
α β
β γ
γ α
tg + tg tg + tg tg = 1 ⇔ xy + yz + zx = 1. MÆt kh¸c:
2 2
2 2
2 2
(x2 + y2 + z2) - (xy + yz + zx) =
[
]
1
( x − y) 2 + ( y − z ) 2 + ( z − x ) 2 ≥ 0
2
⇒ S = x2 + y2 + z2 ≥ xy + yz + zx = 1. Víi x = y = z =
1
th× MinS = 1
3
x , y, z > 0
x
y
z
9
+
+
≤
VD3: Cho
. Chøng minh r»ng: S =
x + yz y + zx z + xy 4
x + y + z = 1
Gi¶i:
xz
β
= tg ;
y
2
§Æt
yz
α
= tg ;
x
2
Do
yz zx
zx xy
xy yz
=x+y+z=1
.
+
.
+.
.
x
y
y
z
z
x
nªn tg
xy
γ
= tg víi α, β, γ ∈
z
2
π
0,
2
α β
β γ
γ α
tg + tg tg + tg tg
=1
2 2
2 2
2 2
α
β γ π α
π α
β γ
β γ
⇔ tg + = cotg ⇔ tg + = tg − ⇔ + = 2 2 2 2
2
2 2
2 2
2 2
⇔
α+β+ γ π
= ⇔ α+β+ γ = π
2
2
S=
2y
2z
3
x
y
z
1 2 x
+
+
=
− 1 +
− 1 +
− 1 +
x + yz y + zx z + xy 2 x + yz y + zx z + xy 2
yz 1 − zx
xy
1−
1
−
1 x − yz y − zx z − xy 3 1
y
x
z
+ 3
+ =
=
+
+
+
+
2 x − yz y + zx z + xy 2 2 1 + yz 1 + zx 1 + xy 2
x
y
z
=
3
1
3 1
(cos + cosβ + cosγ) + = [(cosα + cosβ).1 − (cosα cosβ − sinα + sinβ)] +
2
2 2
2
12
G.NTH
≤
1 1
1
3 3 3 9
2
2
2
(
)
(cos
α
+
cos
β
+
1
)
+
(sin
α
+
sin
β
)
−
cos
α
cos
β
+ 2 = 4 + 2 = 4 (®pcm)
2 2
2
3. C¸c bµi to¸n ®a ra tr¾c nghiÖm
Tríc khi t«i d¹y thö nghiÖm néi dung s¸ng kiÕn cña t«i cho häc sinh cña
2 líp 11A1 vµ 11A2 ë trêng t«i, t«i ®· ra bµi vÒ nhµ cho c¸c em, cho c¸c em
chuÈn bÞ tríc trong thêi gian 2 tuÇn. Víi c¸c bµi tËp sau:
Bµi 1:Cho a2 + b2 = 1. CMR: | 20a3 - 15a + 36b - 48b3| ≤ 13.
Bµi 2:Cho (a-2)2 + (b-1)2 = 5. CMR: 2a + b ≤ 10.
a; b ≥ 0
Bµi 3:Cho
CMR: a4 + b4 ≥ a3 + b3
a + b = 2
Bµi 4:Cho a; b ; c ≥ 1
1
1
1
1
1 1
CMR: a − b − c − ≥ a − b − c −
b
c
a
a
b
c
x; y; z > 0
Bµi 5:Cho 2
2
2
x + y + z + 2 xyz = 1
a) xyz ≤
1
8
3
4
b) xy + yz + zx ≤
c) x2 + y2 + z2 ≥
3
4
d) xy + yz + zx ≤ 2xyz +
e)
CMR:
1
2
1− x
1− y
1− z
+
+
≥ 3
1+ x
1+ y
1+ z
Bµi 6:CMR:
1
1+ a2
+
1
1 + b2
≤
2
∀ a, b ∈ (0, 1]
1 + ab
Bµi 7:CMR: (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9 (ab + bc + ca)
∀ a, b, c > 0
x , y, z > 0
x
y
z
3 3
Bµi 8:Cho
CMR :
+
+
≥
2
2
2
2
1− x 1− y 1− z
xy + yz + zx = 1
x , y, z > 0
x
y
z
3
Bµi 9:Cho
CMR :
+
+
≤
1+ x2
1 + y2
1 + z2 2
x + y + z = xyz
13
G.NTH
x,y,z>0
1
1
1
2x
2y
2z
CMR
:
+
+
≥
+
+
Bµi 10: Cho
1+x2 1+y2 1+z2 1+x2 1+y2 1+z2
xy+yz+zx=1
14
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