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DIN N VT L PH THNG
GII BT NG THC
THI HSG QUC GIA MN TON 2015
http://vatliphothong.vn
Li gii
H
iT
u
a+b+c
Tht vy n lun ng v
ab + bc + ca.
T
ng
Cch 1:
Bt ng thc ny tng ng
[
(
)]
(a b) + (b c) + (c a) (a + b + c) a + b + c
ab + bc + ca 0,
2
[
2
2]
1
(a + b + c) ( a b) + ( b c) + ( c a) 0,
2
a+b+c
Sa = 1
2( b + c)
a+b+c
Sb = 1
2( c + a)2
a+b+c
2
Sc = 1
2( a + b)
Gi s a b c, ta thy
a + c + 4 ac b
a+b+c
S =1
2 =
2 0
b
2( c + a)
2( c + a)
a
+
b
+
c
a
+
b
+
4
ab c
2 =
2 0
Sc = 1
2( a + b)
2( a + b)
Mt khc, ta c
ac
a
, do
bc
b
H
iT
u
)
(
a+b+c
a+b+c
2
a2 1
2 + b 1 (
)2 0,
2( c + a)
2 b+ c
b2
a2
a2 + b2 (a + b + c)
)2 .
2 + (
2( c + a)
2 b+ c
T
ng
)
(
2
2
a
b
a2
b2
(a + b + c)
2 + (
)2 (a + b + c) 2 (c + a) + 2 (b + c)
2( c + a)
2 b+ c
a2 + b2
a2 b
b2 a
+
+
2
2 (c + a) 2 (b + c)
2
2
a2 b b2 a
a +b
+
+
2
2a
2b
a2 + b2
=
+ ab
2
a2 + b2 a2 + b2
+
2
2
= a 2 + b2 .
=
Ta c,
H
iT
u
Cch 3:
Bt ng thc cn chng minh tng ng
ab( a b)2
0.
a+b
2ab
2bc
2ca
T suy ra ab
. Tng t ta cng c bc
, ca
.
a+b
b+c
c+a
S dng cc bt ng thc ny v bt ng thc Cauchy Schwarz ta c
(a + b + c)( ab + bc + ca) + a2 + b2 + c2
)
(
ab
bc
ca
+
+
+ a2 + b2 + c2
2(a + b + c)
a+b b+c c+a
(
)
1
1
1
= 2(ab + bc + ca) + 2abc
+
+
+ a2 + b2 + c2
a+b b+c c+a
(
)
9
2(ab + bc + ca) + 2abc
+ a2 + b2 + c2
a+b+b+c+c+a
9abc
+ a2 + b2 + c2 .
= 2(ab + bc + ca) +
a+b+c
T
ng
2ab
ab
=
a+b
2ab
ab
=
a+b
ab( a b)2
0.
a+b
3
2ab
2bc
2ca
. Tng t ta cng c bc
, ca
.
a+b
b+c
c+a
S dng cc bt ng thc ny ta c
(a + b + c)( ab + bc + ca) + a2 + b2 + c2
(
)
ab
bc
ca
2(a + b + c)
+
+
+ a2 + b2 + c2
a+b b+c c+a
(
)
1
1
1
= 2(ab + bc + ca) + 2abc
+
+
+ a2 + b2 + c2 .
a+b b+c c+a
T suy ra
ab
(
2abc
rng
1
1
1
+
+
a+b b+c c+a
2abc
+ 2a2 = 2a
b+c
2 (ab + bc + ca)
hay l chng minh
(
)
+ 2 a2 + b2 + c2 (a + b + c)2 .
H
iT
u
tng ng
bc
+a
b+c
2a (ab + bc + ca)
.
b+c
a
b
c
+
+
b+c c+a a+b
(a + b + c)2 ,
a
b
c
(a + b + c)2
+
+
.
b+c c+a a+b
2ab + 2bc + 2ca
Bt ng thc ny hin nhin ng theo Cauchy Schwarz
T
ng
b
c
a2
b2
c2
a
+
+
=
+
+
b+c c+a a+b
ab + ac bc + ba ca + cb
(a + b + c)2
.
2ab + 2bc + 2ca
Cch 5:
D thy rng khi abc = 0 th bt ng thc hin nhin ng.
Xt abc > 0.
V bt ng thc hon ton thun nht, nn ta c th chun ha abc = 1.
Bt ng thc cn chng minh tng ng
(a + b + c)( ab + bc + ca)
(
)
a+ b+ c
= ab (a + b) + bc (b + c) + ca (c + a) + abc
3
2ab + 2bc + 2ca + abc 3
abc
= 2 (ab + bc + ca) + 3
= 2 (ab + bc + ca) + 2abc + 1.
4
Do , ta cn chng minh
a2 + b2 + c2 + 2abc + 1 2(ab + bc + ca).
Theo nguyn l Dirichlet th trong 3 s dng a, b, c lun tn ti hai s nm cng pha so vi 1.
Gi s hai s l a v b. Khi ta c 2c(a 1)(b 1) 0 hay tng ng
2abc + 2c 2(bc + ca).
T , chng minh s hon tt nu ta ch ra c
a2 + b2 + c2 + 2abc + 1 2ab + 2abc + 2c,
(a b)2 + (c 1)2 0.
H
iT
u
hay
T
ng
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