221_bat dang thuc
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hsmath.net gii thiu:
BT NG THC1. K thut thm bt S dng: A A B B gi c cc b phn vi nhau Cc v d: Bi 1: Chng minh rng vi mi a,b,c>0, ta c:a2 b c b2 c a c2 a b a b c 2
A B to ra cc b phn mi hai v ca bt ng thc m c th nh B
Phn tch: - BT ng bc nht - Vai tr a,b,c ging nhau - D on du bng xy ra khi a=b=c - Biu thc thm vo l bc nht Hng dn:a2 b c b c 4 2a
Bi 2: Chng minh rng vi mi a,b,c>0, ta c:a3 b 2c b3 c 2a c3 a 2b a2 b2 3 c2
Phn tch: - BT ng bc hai - Vai tr a,b,c ging nhau - D on du bng xy ra khi a=b=c - Biu thc thm vo l bc hai Hng dn:a3 a(b 2c) 2 2 a b 2c 9 3 ab bc ca a 2 b 2 c 2
Bi 3: Chng minh rng vi mi a,b,c>0, ta c:(1 a3 )(1 b3 )(1 c3 ) (1 ab 2 )(1 bc 2 )(1 ca 2 )
Phn tch: - D on du bng xy ra khi a=b=c Hng dn:(1 a3 )(1 b3 )(1 b3 ) 13
a3b3c3
3
1 ab2
3
Bi tp: Cho a,b,c>0. Chng minh cc bt ng thc sau: 1)a5 b2 b5 c2 c5 a2 a3 b3 c3
2)
a3 (b c) 2 a3 b (c a )
b3 (c a ) 2 b3 c(a b)
c3 ( a b) 2 c3 a (b c)
a b c 4 a b c 2
3)
hsmath.net gii thiu:
4)
a4 bc 2
b4 ca 2
c4 ab 2
a b c
5)
a
2
a3 ab b 2
b
2
b3 bc c 2
c
2
c3 ca a 2
a b c 3
Cho a,b,c>0 v a+b+c=1. Chng minh cc bt ng thc sau: 6)a5 b4 b5 c4 c5 a4 1
7)
a b c
b c a
c a b
3 3 2
Chng minh rng trong tam gic nhn ABC, ta c: 8)1 cosA 1 cosB 1 6 cosB 1 2 cos2B 1 cosBcosC 1 2 cos2B 6 5 27 2
9)
1 2 cos2A 1 cosAcosB
10)
1 cosA+cosB+cosC cosCcosA
2. K thut san s Xc nh: i lng ln, i lng b v chn cch san s ph hp Cc v d: Bi 1: Chng minh rng vi mi x,y>0 v x+y=1, ta c:1 x2
y
2
1 xy
4 xy 7
Phn tch: - Vai tr x,y ging nhau - D on du bng xy ra khi x=y= - i lng ln: Hng dn:1 x2
1 2
1 1 ;4xy ; i lng b: 2 xy x y2
y2
2
1 xy2
4 xy 2
1 x2
y
2
1 2 xy 1 (x
1 4 xy y)2 7
4 xy
1 4 xy
(1 1) x y 2 2 xy
1 .4 xy 4 xy
Bi 2: Chng minh rng trong tam gic nhn ABC, ta c:1 cosA 1 cosB 1 15 cosA+cosB+cosC cosB 2
hsmath.net gii thiu: Phn tch: - Vai tr A,B,C ging nhau - D on du bng xy ra khi A=B=C=600 - i lng ln: Hng dn:1 cosA 1 cosB 1 ; i lng b: cosA+cosB+cosC cosC
1 1 1 cosA+cosB+cosC cosA cosB cosC 1 1 1 4cosA +4cosB 4cosC cosA cosB cosC 9 15 -3(cosA cosB cosC) 4 4 4 2 2
Bi tp: Cho a,b,c>0. Chng minh cc bt ng thc sau: 1) 2)2(a3 b3 c3 ) abca4 b4 c4
9(a b c)3 a 2 b2 c2a 2b 2 b2c2
33c2a2 a 3b b 3 c c 3 a ab3 bc 3 ca 3
3) Cho x,y>0 v x+y=1. Tm GTNN ca biu thc:P 3 x2
y
2
2 xy
4 xy; Q
1 x2
y
2
1 xy
3. K thut nhm i xng Bt ng thc dng i xng (vai tr ca cc bin l nh nhau). Khi chng ta c th nh gi mt b phn ca v ny vi b phn tng ng ca v kia. Tng t, suy ra cc kt qu i vi cc b phn cn li v thu c bt ng thc cn chng minh. Cc v d: Bi 1: Chng minh rng vi mi a,b,c>0, ta c: Phn tch: - BT ng bc nht - Vai tr a,b,c - D on du bng xy ra khi a=b=c Hng dn:bc a ca b 2 bc ca . a b 2b
bc a
ca b
ab c
a b c
Bi 2: Chng minh rng trong tam gic ABC, ta c:sin A sin B sin C cos A 2 cos B 2 cos C 2
Phn tch: - Vai tr A,B,C ging nhau - D on du bng xy ra khi A=B=C=600 Hng dn:
hsmath.net gii thiu:sin A 4sin sin B A B 2 2(sin A sin B) 2 cos C 2
Bi 3: Chng minh rng trong tam gic ABC, ta c:sin A sin B sin C sin A 3B B 3C C 3A sin sin 4 4 4
Hng dn:sin A 3B 4 1 A B sin sin B 2 24
1 3 sin A sin B 4 4
1 4 .4. sin A sin 3 B 4
sin A sin 3 B
Bi tp: Cho a,b,c>0. Chng minh cc bt ng thc sau: 1)a2 b2 b2 c2 c2 a2 a c b a c b
2)
ab a b
bc b c
ca c a
a b c 2
Chng minh rng trong tam gic ABC, ta c: 3) sin2 A sin2B sin2C sin A sin B sin C 4) cos A cos BcosC sin sin sin 5)1 sin 2 An
A 2
B 2
C 21 1 B 2cos
1 sin 2 An
1 sin 2 An
1 cos2n
A 2cos
cos2A 2n
cos2B 2n
C 2cos C 2
6)
sin A
sin B
sin C
4. K thut ng bc ho S dng gi thit bin i BT v dng ng bc chng minh. Cc v d: Bi 1: Chng minh rng vi mi a,b>0 v a+b=1, ta c:ab(a 2 b2 )
Phn tch: - BT khng ng bc - Vai tr a,b ging nhau - D on du bng xy ra khi a=b - S dng gii thit ng bc ho Hng dn:
1 8
hsmath.net gii thiu:ab(a 2 b2 ) 1 ( a b) 4 8 8ab(a 2 b 2 ) 0 0
( a b) 4 ( a b) 4
Bi 2: vi mi a,b,c>0 v a+b+c=1. Chng minh rng :a2 b2 c2 2 3abc 1
Phn tch: - BT khng ng bc - Vai tr a,b ging nhau - D on du bng xy ra khi a=b=c= - S dng gii thit ng bc ho Hng dn:a2 b2 c2 2 3abc(a b c) 3abc(a b c) (ab bc ca )
1 3
(a b c ) 2
3abc(a b c) (ab bc ca ) 2
Bi tp: 1) Cho a,b,c>0, tho iu kin:a2 3
a
2
3
a2
2
3
32
Chng minh rng : a b c 2) Cho a,b>0, tho iu kin:a b 2 Chng minh rng : 2 a2 b216a 16b 16c
2
a
2
3
b
2
3
c
2
3
a 3 b3 2a 2b 2c
a4
b4
3) Chng minh rng vi mi a,b,c: a+b+c=0, ta c:
5. K thut chun ho S dng tnh cht ng bc ca BT chun ho. Vic chn i tng chun ho l rt quan trng. Cc v d: Bi 1: Cho a,b,c>0. Chng minh rng:a(b c) (b c)2 a 2 b (c a ) (c a ) 2 b 2 c (a b) (a b) 2 c 2 6 5
Phn tch: - BT ng bc - Vai tr a,b,c ging nhau - D on du bng xy ra khi a=b=c - Chun ho: a + b + c = 1 Hng dn:a(1 a) 1 2a 2a 2 b(1 b) 1 2b 2b22
c(1 c) 1 2c 2c 2a 1 42
Theo Csi: 2a(1-a)
2a 1 a 2
=
hsmath.net gii thiu: => 1- 2a + 2a = 1 - 2a (1- a) =>a(1 a) 1 2a 2a 2 4a(1 a) (1 a)( a 3)2
14 a
a 1 4
2
=
1 a a 3 43 a 3
>0
a 3
4 1
=> VT
4 (1
3 a 3
) (1
3 b 3
) (1
3 c 3
) = 4 3 3(
1 a 3
1 b 3
1 c 3
6 5
Bi 2: Cho a, b, c>0. Chng minh rng: 6(a + b + c) (a2 + b2 + c2) 27abc + 10 (a2+b2+c2)3/2 (1) Phn tch: - BT ng bc - Vai tr a,b,c ging nhau - D on du bng xy ra khi a=b=c - Chun ho: a2 + b2 + c2 =9 Hng dn: (1) 2(a + b + c) - abc VT2 10
VT = 2(a+b+c) - abc = 2a - abc + 2(b+c) = a(2-bc) + 2(b+c) [a2 + (b+c)2] [(2- bc)2 + 4] 33
G/s: a b c do a2 + b2 + c2 = 9 => a2 t t = bc do bc Nn VT2b2 2 c2 9 a2 2
(9+2bc) [(2-bc)2 + 4] = (9 + 2t) [(2-t)2 + 4] = f(t) vi -3 t 3 max f(t) = 100 => VT 10 pcm
Kho st f(t) => f(t)
1) (a+b) (b+c) (c+a) + abc 2)
1 (a + b + c)3; a, b, c > 0 32;
8abc a 2 b2 c 2 + ab bc ca (a b)(b c)(c a)
a, b, c > 0a 2 b2 c 2 ab bc ca
3) a, b, c > 0:
( a b c) 2 a 2 b2 c 2
1 a 3 b3 c 3 2 abc
25
4) a, b, c > 0: (a + b + c) (
1 a b
1 b c
1 c a
)+
4abc (a b)(b c)(c a )
6. K thut lng gic ho K thut lng gic ho vi mc ch thay i hnh thc ca bi ton chng minh mt BT i s thnh vic chng minh BT lng gic. K thut ny c xc nh thng qua min gi tr ca cc bin, cc cng thc lng gic v cc ng thc lng gic lin quan. Cc v d: Bi 1: Chng minh rng:a 1 b2 b 1 a2 3(ab (1 a 2 )(1 b 2 ) 2
hsmath.net gii thiu: Phn tch: - K: 1 a, b 1 - Cng thc lng gic lin quan sin 2 - Lng gic ho Hng dn: t:a sin b sin
cos2
1
;
,
0;
VT= 2 sin(abc
3
)
2
Bi 2: Cho x,y,z>0; zy+yz+zx=1. Chng minh rng:(1 a )(1 b)(1 c) 1
Phn tch: - ng thc lng gic lin quan tg tg - Lng gic ho Hng dn:
A 2
B B C tg tg 2 2 2
tg
C A tg 1 2 2
A B C ; ABC l tam gic nhn ;b t g ;c t g 2 2 2 1 3 3 VT tgA tgB tgC 2 2 1 1 1 Bi 3: Cho a,b,c l cc s dng tha mn: 6 a 2b 3c
t:
a
tg
Chng minh rng:
a b c a 36bc b 9ca c 4ab
1 27
Hng dn:1 1 1 36bc 9ca 4ab 1 1 1 a b c 36bc A 9ca B t cotg 2 , cotg 2 , 0 A, B a 2 b 2 bc ca ab bc ca ab 3 2 6 3 2 T gi thit ta c: 6 a b c a b c A B cotg cotg ab C 2 2 tg A B 2 cotg Suy ra, A B c 2 2 cotg cotg 1 2 2 VT
vi A,B,C l ba gc ca mt tam gic Vy VT1 1 A B C 1 cotg 2 1 cotg 2 1 cotg 2 2 2 2 1
hsmath.net gii thiu:sin 2 A 2B C sin sin 2 2 22
1 A-B A+B cos cos 4 2 2 sin C 2 1 C 1 sin 4 2 1 82
2
sin C 2
C 2
1 A-B C cos sin 4 2 2 1 C 1 sin 8 2
sin C 2
1 sin
C C 2sin 2 2
1 sin
1 sin 3
C 2
2sin
C 2
3
1 27
Bi tp: 1) Cho 0