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TNG HI TUN

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BT NG THC
QUA CC THI CHN HSG MN TON
CA CC TRNG, CC TNH TRN C NC
NM HC 2014 - 2015

H Ni - 2015

bi

Bi 1. Cho x, y, z l cc s thc dng tha mn 3 (x4 + y 4 + z 4 ) 7 (x2 + y 2 + z 2 ) + 12 = 0. Tm


gi tr nh nht ca biu thc
x2
y2
z2
P =
+
+
.
y + 2z z + 2x x + 2y
Chn HSG Quc gia, Yn Bi, 2014 - 2015
Bi 2. Cho 2014 s thc dng a1 , a2 , ..., a2014 c tng bng 2014. Chng minh rng
a20
a20
a20
1
2
2014
+
+
...
+
2014.
11
11
a2
a3
a11
1
Chn HSG Quc gia, Cn Th, 2014 - 2015
Bi 3. Tm hng s k ln nht vi mi a, b, c khng m tha mn a + b + c = 1 th bt ng thc
sau ng
a
b
c
1
2 +
2 +
2 .
2
1 + 9bc + k(b c)
1 + 9ca + k(c a)
1 + 9ab + k(a b)
Chn HSG Quc gia, Hi Phng, 2014 - 2015
Bi 4. Cho cc s thc x, y, z thay i tha mn 4x + 4y + 4z = 1. Tm gi tr ln nht ca
S = 2x+2y + 2y+2z + 2z+2x 2x+y+z
Chn HSG Quc gia, Hi Dng, 2014 - 2015
Bi 5. Cho cc s x, y tha mn: 0 < x 1, 0 < y 1. Tm gi tr nh nht ca biu thc
F =

x5 + y + 4 y 4 2y 3 + x
+
.
x
y2
Chn HSG Quc gia, C Mau, 2014 - 2015

Bi 6. Cho a, b, c l cc s thc dng tha mn ab + bc + ca = 1. Chng minh rng


3

a3
b3
c3
(a + b + c)
+
+

.
2
2
2
1 + 9b ac 1 + 9c ba 1 + 9a cb
18
Chn HSG Quc gia, chuyn Quc hc Hu, 2014 - 2015
Bi 7. Cho a, b, c l cc s khng m, khng c hai s no trong cc s ng thi bng khng.
Tm gi tr nh nht ca biu thc:
P =

a(b + c) b(c + a) c(a + b)


+ 2
+ 2
.
a2 + bc
b + ca
c + ab
Chn HSG Quc gia, Thanh Ha, 2014 - 2015

Bi 8. Chng minh rng vi mi s thc dng a, b, c ta c


b(a + c)
c(a + b)
6
a(b + c)
+
+

.
(b + c)2 + a2 (a + c)2 + b2 (a + b)2 + c2
5
Chn HSG Quc gia, Thi Bnh, 2014 - 2015
1

Bi 9. Cho x, y, z l cc s khng m. Chng minh rng


xyz + x2 + y 2 + z 2 + 5 3 (x + y + z) .
Chn HSG Quc gia, Chuyn L Qu n - Ninh Thun, 2014 - 2015
Bi 10. Cho a, b, c l cc s thc dng. Chng minh rng
(a + b c)2
(a + c b)2
(c + b a)2
3
+
+
.
2
2
2
2
2
2
(a + b) + c
(a + c) + b
(c + b) + a
5
Chn HSG Quc gia, k Lk, 2014 - 2015
Bi 11. Chng minh bt ng thc sau
3(x2 x + 1)(y 2 y + 1) 2(x2 y 2 xy + 1), x, y R.
Du "=" xy ra khi no?
Chn HSG Quc gia, Qung Tr, 2014 - 2015
Bi 12. Cho x, y, z l cc s thc khng m v i mt phn bit. Chng minh rng
x+y
y+z
z+x
9
.
2 +
2 +
2
x+y+z
(x y)
(y z)
(z x)
Chn HSG Quc gia, Chuyn H S phm H Ni, 2014 - 2015
Bi 13. Cho a, b, c l cc s thc dng tha mn a + b + c = 3. Chng minh rng
)
(
1 1 1
3
3
3
+ +
3(ab + bc + ca).
a +b +c +2
a b c
Chn HSG quc gia, Lm ng, 2014 - 2015
Bi 14. Cho ba s khng m a, b, c. Chng minh rng:

5a2 + 4bc + 5b2 + 4ca + 5c2 + 4ab 3(a2 + b2 + c2 ) + 2( ab + bc + ca).


Chn HSG quc gia, Qung Nam, 2014 - 2015

Bi 15. Cho ba s thc dng x, y, z tha mn 2 xy + xz = 1. Chng minh rng:


3yz 4zx 5xy
+
+
4.
x
y
z
Chn HSG quc gia, Tuyn Quang, 2014 - 2015
Bi 16. Cho x, y, z l cc s thc dng tha mn x + y + z = xyz. Chng minh rng
2
1
1
9

+
+
.
2
2
2
4
1+x
1+z
1+y
Chn HSG quc gia, Thi Nguyn, 2014 - 2015

Bi 17. Cho cc s thc khng m x, y, z tha mn: x2 + y 2 + z 2 = 2. Tm gi tr ln nht ca


M=

x2
y+z
1
+
+
.
x2 + yz + x + 1 z + y + x + 1 xyz + 3
Chn HSG Quc gia, Chuyn Hng Vng - Ph Th, 2014 - 2015

Bi 18. Cho a, b, c l cc s thc dng tha mn a + b + c = 6. Chng minh rng khi ta c


a2 + bc b2 + ca c2 + ab
+
+
a2 + b2 + c2 .
b
c
a
Chn HSG tnh, Hi Phng, 2014 - 2015
Bi 19. Cho a, b l 2 s tha mn iu kin: a2 + b2 + 9 = 6a + 2b. Chng minh
4b 3a.
Chn HSG tnh Bnh Thun, 2014 - 2015
Bi 20. Cho ba s dng a, b, c tha mn a + b + c = 3. Tm gi tr nh nht ca biu thc
P = 7(a4 + b4 + c4 ) +

ab + bc + ca
.
+ b2 c + c2 a

a2 b

Chn HSG tnh B Ra Vng Tu, 2014 - 2015


Bi 21. Cho a, b v c l cc s thc dng. Tm gi tr nh nht ca biu thc:
P =

a + 3c
4b
8c
+

.
a + 2b + c a + b + 2c a + b + 3c
Chn HSG tnh Kin Giang, 2014 - 2015

Bi 22. Cho a, b, c l ba s thc dng tha mn abc = 1. Chng minh rng:


a3

1
1
1
+ 3
+ 3
1.
3
3
+ b + 1 b + c + 1 c + a3 + 1
Chn HSG tnh Long An, 2014 - 2015

Bi 23. Cho cc s thc a, b, c 1 tha mn a + b + c = 6. Chng minh rng:


(a2 + 2)(b2 + 2)(c2 + 2) 216.
Chn HSG tnh Vnh Phc, 2014 - 2015
Bi 24. Cho a, b, c l cc s thc dng. Tm gi tr ln nht ca biu thc

8a + 3b + 4( ab + bc + 3 abc)
P =
.
1 + (a + b + c)2
Chn HSG tnh, Thanh Ha, 2014 - 2015
Bi 25. Cho x, y, z l cc s thc dng tha mn xy + yz + zx = 2xyz. Chng minh rng:

x
y
z
+
+
1.
2
2
2
2
2
2
2y z + xyz
2z x + xyz
2x y + xyz
Chn HSG tnh, Gia Lai, 2014 - 2015
3

Bi 26. Cho x, y, z l cc s thc dng tha mn iu kin x 1, y 2 v x + y + z = 6. Chng


minh rng
(x + 1) (y + 1) (z + 1) 4xyz.
thi chuyn h lp 10, THPT Chuyn S phm, 2014 - 2015
Bi 27. Cho cc s thc dng a, b, c tha mn a + b + c = 1. Chng minh rng
(
)
1+a 1+b 1+c
a b c
+
+
2
+ +
.
1a 1b 1c
b c a
Chn i tuyn Olympic Ton lp 10 vng 1, Chuyn Nguyn Du, 2014 - 2015
Bi 28. Cho a, b, c, d l cc s thc dng tha mn a + b + c + d = 4. Chng minh rng

(a + b)2
(b + c)2
(c + d)2
(d + a)2
P =
+
+
+
16.
a2 ab + b2
b2 bc + c2
c2 cd + d2
d2 ad + a2
thi kho st i tuyn lp 10 vng 2, Chuyn KHTN, 2014 - 2015
Bi 29. Cho x, y, z l cc s thc dng. Chng
)
(
y)(
x (
1+
1+
1+
y
z

minh:
z)
x+y+z
2+2
.
3 xyz
x

Chn i tuyn d thi Olympic 30-4 lp 10, tnh Bnh Thun, 2014 - 2015

Li gii

Bi 1. Cho x, y, z l cc s thc dng tha mn 3 (x4 + y 4 + z 4 ) 7 (x2 + y 2 + z 2 ) + 12 = 0. Tm


gi tr nh nht ca biu thc
P =

x2
y2
z2
+
+
.
y + 2z z + 2x x + 2y
Chn HSG Quc gia, Yn Bi, 2014 - 2015
Li gii
2

S dng bt ng thc Cauchy Schwarz, ta c 3(x4 + y 4 + z 4 ) (x2 + y 2 + z 2 ) , do


(
)2
0 x2 + y 2 + z 2 7(x2 + y 2 + z 2 ) + 12.
T suy ra x2 + y 2 + z 2 3. S dng bt ng thc Cauchy Schwarz, ta li c
x2
y2
z2
+
+
y + 2z z + 2x x + 2y
x4
y4
z4
= 2
+
+
x y + 2zx2 y 2 z + 2xy 2 z 2 x + 2yz 2
2
(x2 + y 2 + z 2 )
2
.
x y + y 2 z + z 2 x + 2 (xy 2 + yz 2 + zx2 )

P =

Tip tc s dng bt ng thc Cauchy Schwarz v kt hp vi bt ng thc quen thuc


(a + b + c)2
ab + bc + ca
, ta c
3

x2 y + y 2 z + z 2 x (x2 + y 2 + z 2 ) (x2 y 2 + y 2 z 2 + z 2 x2 )

(x2 + y 2 + z 2 )2
2
2
2
(x + y + z )
3

(
) (x2 + y 2 + z 2 )
= x2 + y 2 + z 2
.
3
Hon ton tng t, ta chng minh c
(

2 xy 2 + yz 2 + zx

)
2

2 x2 + y 2 + z

)
2

(x2 + y 2 + z 2 )
.
3

T suy ra
2

(x2 + y 2 + z 2 )

P
(x2 + y 2 + z 2 )
2
2
2
3 (x + y + z )
3

x2 + y 2 + z 2
=
3
1.
ng thc xy ra khi v ch khi x = y = z = 1 nn gi tr nh nht ca P l 1. 

Bi 2. Cho 2014 s thc dng a1 , a2 , ..., a2014 c tng bng 2014. Chng minh rng
a20
a20
a20
1
2
2014
+
+
...
+
2014.
11
11
a11
a
a
2
3
1
Chn HSG Quc gia, Cn Th, 2014 - 2015
Li gii
S dng bt ng thc AM GM cho 20 s dng, ta c

20
20
a1
20 a1
8
+
11

a
+
8

20

a11
2
2 1 = 20 a1 .
11
a11
a
2
2
Tng t vi 2013 s hng cn li, sau cng v vi v, v vi ch

2014

ai = 2014 ta thu ngay

i=1

c iu phi chng minh.


ng thc xy ra khi v ch khi tt c cc bin bng nhau v bng 1. 
Bi 3. Tm hng s k ln nht vi mi a, b, c khng m tha mn a + b + c = 1 th bt ng thc
sau ng
a
b
c
1
2 +
2 +
2 .
2
1 + 9bc + k(b c)
1 + 9ca + k(c a)
1 + 9ab + k(a b)
Chn HSG Quc gia, Hi Phng, 2014 - 2015
Li gii
1
v c = 0 ta c k 4 v ta s chng minh kmax = 4. Tht vy, vi k = 4 bt
2
ng thc cn chng minh tr thnh
Cho a = b =

a
b
c
1
2 +
2 +
2 .
2
1 + 9bc + 4(b c)
1 + 9ca + 4(c a)
1 + 9ab + 4(a b)
K hiu v tri l A, s dng bt ng thc Cauchy Schwarz ta c
(a + b + c)2
)
A (
a + 9abc + 4a(b c)2
1
=
.
2
1 + 27abc + 4a(b c) + 4b(c a)2 + 4c(a b)2
Do , ta quy bi ton v chng minh
1 + 27abc + 4a(b c)2 + 4b(c a)2 + 4c(a b)2 2.
Hay tng ng
4ab(a + b) + 4bc(b + c) + 4ca(c + a) + 3abc 1.
ng bc ha bt ng thc ny, ta cn chng minh
4ab(a + b) + 4bc(b + c) + 4ca(c + a) + 3abc (a + b + c)3 ,
hay tng ng
a3 + b3 + c3 + 3abc ab(a + b) + bc(b + c) + ca(c + a).
6

y chnh l bt ng thc Schur bc 3, bi ton chng minh xong.


1
ng thc xy ra khi v ch khi a = b = , c = 0 hoc cc hon v. 
2
Bi 4. Cho cc s thc x, y, z thay i tha mn 4x + 4y + 4z = 1. Tm gi tr ln nht ca
S = 2x+2y + 2y+2z + 2z+2x 2x+y+z
Chn HSG Quc gia, Hi Dng, 2014 - 2015
Li gii
t a = 2x , b = 2y , c = 2z th ta c a, b, c > 0 v a2 + b2 + c2 = 1. Khi ta cn tm gi tr
ln nht ca biu thc
S = ab2 + bc2 + ca2 abc.
Khng mt tnh tng qut, ta c th gi s b l s nm gia hai s a v c.
Khi ta c a(a b)(b c) 0, tng ng
a2 b + abc ca2 + ab2 .
S dng nh gi ny, kt hp vi bt ng thc AM GM b ba s, ta c
S a2 b + bc2
1
= 2b2 (a2 + c2 ) (a2 + c2 )
2
(
)3
2b2 + (a2 + c2 ) + (a2 + c2 )
1

3
2

2 3
=
.
9

1
2 3
ng thc xy ra khi v ch khi a = b = c = nn gi tr ln nht ca S l
.
9
3
Bi 5. Cho cc s x, y tha mn: 0 < x 1, 0 < y 1. Tm gi tr nh nht ca biu thc
F =

x5 + y + 4 y 4 2y 3 + x
+
.
x
y2
Chn HSG Quc gia, C Mau, 2014 - 2015
Li gii

S dng bt ng thc AM GM v ch

y 1, ta c

x5 + y + 4 y 4 2y 3 + x
+
x
y2
y 4
x
= x4 + + + y 2 2y + 2
x x
y
(
)
) (
1
1
1
1
x
y
4
= x + + + +
+ 2 + (y 1)2 1
+
x x x x
x y
2
5+ 1
y

F =

5+21
= 6.
7

ng thc xy ra khi v ch khi x = y = 1 nn gi tr nh nht ca F l 6. 


Bi 6. Cho a, b, c l cc s thc dng tha mn ab + bc + ca = 1. Chng minh rng
3

a3
b3
c3
(a + b + c)
+
+

.
1 + 9b2 ac 1 + 9c2 ba 1 + 9a2 cb
18
Chn HSG Quc gia, chuyn Quc hc Hu, 2014 - 2015
Li gii
S dng bt ng thc Holder, ta c
V T (1 + 9b2 ac + 1 + 9c2 ba + 1 + 9a2 cb) (1 + 1 + 1) (a + b + c)3 .
Do ta cn chng minh rng
1 + 9b2 ac + 1 + 9c2 ba + 1 + 9a2 cb 6,
tng ng
3abc(a + b + c) 1 = (ab + bc + ca)2 .
Bt ng thc cui lun ng nn php chng minh hon tt.
1
ng thc xy ra khi v ch khi a = b = c = . 
3
Bi 7. Cho a, b, c l cc s khng m, khng c hai s no trong cc s ng thi bng khng.
Tm gi tr nh nht ca biu thc:
P =

a(b + c) b(c + a) c(a + b)


+ 2
+ 2
.
a2 + bc
b + ca
c + ab
Chn HSG Quc gia, Thanh Ha, 2014 - 2015
Li gii

Bi ny mnh khng gii c, mi cc bn tham kho 2 li gii sau y:


Cch 1 (Nguyn Vn Qu - quykhtn-qa1):
Khng mt tnh tng qut, gi s rng a b c 0, khi
a(b + c)
a(b + c)
b+c
b
2
=
,
2
a + bc
a + ac
a+c
a
v

c(a + b)
c(a + b)
c
2
= .
2
c + ab
b + ab
b

T ,
a(b + c) b(c + a) c(a + b)
b
ab
c
b2 + ca
ab
+
+

+
+
=
+ 2
2.
2
2
2
2
a + bc
b + ca
c + ab
a b + ca b
ab
b + ca
ng thc xy ra khi a = b, c = 0 hoc cc hon v nn gi tr nh nht ca P l 2. 
Cch 2 (V Quc B Cn):
Khng mt tnh tng qut, gi s a b c. Khi , ta c
(a b)(c a) (a b)(b c)
a(b + c) b(c + a)
+ 2
2=
+
2
a + bc
b + ca
a2 + bc
b2 + ca
2
(a b) (c2 2ac 2bc + ab)
=
(a2 + bc)(b2 + ca)
(a b)2 (c2 + ab)
2c(a + b)(a b)2
= 2

.
(a + bc)(b2 + ca) (a2 + bc)(b2 + ca)
8

Bt ng thc cn chng minh tr thnh


c(a + b)
2c(a + b)(a b)2
(a b)2 (c2 + ab)
+ 2
2
.
(a2 + bc)(b2 + ca)
c + ab
(a + bc)(b2 + ca)
n y, s dng bt ng thc AM GM ta c

2(a b) c(a + b)
(a b)2 (c2 + ab)
c(a + b)
+ 2

.
(a2 + bc)(b2 + ca)
c + ab
(a2 + bc)(b2 + ca)

T , bi ton c a v chng minh


(a2 + bc)(b2 + ca) c(a + b)(a b)2 ,
hin nhin ng do ta c a2 + bc a2 (a b)2 v b2 + ca c(a + b).
Php chng minh c hon tt. Du ng thc xy ra khi v ch khi a = b, c = 0 (v cc hon v
tng ng). 
Bi 8. Chng minh rng vi mi s thc dng a, b, c ta c
a(b + c)
b(a + c)
c(a + b)
6
+
+

.
(b + c)2 + a2 (a + c)2 + b2 (a + b)2 + c2
5
Chn HSG Quc gia, Thi Bnh, 2014 - 2015
Li gii
Cch 1: Do tnh thun nht nn ta c th chun ha cho a + b + c = 3, khi ta c
a(b + c)
a(3 a)
=
2
2
(b + c) + a
(3 a)2 + a2
9a + 1 9(a 1)2 (2a + 1)
]
=
[
25
25 (3 a)2 + a2
9a + 1

.
25
Tng t vi hai biu thc cn li, sau cng v vi v v ch a + b + c = 3 ta thu ngay c
iu phi chng minh.
ng thc xy ra khi v ch khi a = b = c. 
(b + c)2
Cch 2: S dng bt ng thc AM GM , ta c a2 +
a(b + c), t
4
a(b + c)
a(b + c)

2
2
(b + c) + a
3(b + c)2
+ a(b + c)
4
3(b + c)2
4
=1
2
3(b + c)
+ a(b + c)
4
3(b + c)2
=1
.
3(b + c)2 + 4a(b + c)
Bi ton a v chng minh
(b + c)2
(c + a)2
(a + b)2
3
+
+
.
2
2
2
5
3(b + c) + 4a(b + c) 3(c + a) + 4b(c + a) 3(a + b) + 4c(a + b)
9

S dng bt ng thc Cauchy Schwarz ta c


4(a + b + c)2
VT
.
3(b + c)2 + 4a(b + c) + 3(c + a)2 + 4b(c + a) + 3(a + b)2 + 4c(a + b)
T bi ton s c chng minh nu ta ch ra c
4(a + b + c)2
3
.
2
2
2
5
3(b + c) + 4a(b + c) + 3(c + a) + 4b(c + a) + 3(a + b) + 4c(a + b)
Tht vy, sau khi quy ng, kh mu v rt gn, th bt ng thc trn tng ng vi
2(a2 + b2 + c2 ) 2(ab + bc + ca).
Hin nhin ng. Bi ton c chng minh xong.
ng thc xy ra khi v ch khi a = b = c. 
Bi 9. Cho x, y, z l cc s khng m. Chng minh rng
xyz + x2 + y 2 + z 2 + 5 3 (x + y + z) .
Chn HSG Quc gia, Chuyn L Qu n - Ninh Thun, 2014 - 2015
Li gii
Theo nguyn l Dirichlet th trong ba s x, y, z lun tn ti hai s nm cng pha so vi 1, khng
mt tnh tng qut ta c th gi s rng hai s l x v y. Khi z(x 1)(y 1) 0, hay tng
ng
xyz xz + yz z.
Cch 1: S dng nh gi ny, ta quy bi ton v chng minh
f (z) = z 2 + (x + y 4) z + x2 + y 2 + 5 3x 3y 0.
y l mt hm bc hai theo z vi h s ca z 2 dng, mt khc
= (x + y 4)2 4(x2 + y 2 3x 3y + 5)
= 3x2 3y 2 + 2xy + 4x + 4y 4
= (x y)2 2(x 1)2 2(y 1)2
0.
Nn t suy ra f (z) 0, z.
Bi ton c chng minh xong.
ng thc xy ra khi v ch khi x = y = z = 1. 
Cch 2: S dng bt ng thc AM GM ta c
(x + y + z)2 + 9
,
3(x + y + z)
2
nn ta quy bi ton v chng minh bt ng thc mnh hn l
2xyz + 2(x2 + y 2 + z 2 ) + 10 (x + y + z)2 + 9,
hay tng ng
x2 + y 2 + z 2 + 2xyz + 1 2(xy + yz + zx).
10

S dng xyz xz + yz z th ta cn phi chng minh


x2 + y 2 + z 2 + 2(xz + yz z) + 1 2xy + 2yz + 2zx,
hay
(x y)2 + (z 1)2 0.
Bt ng thc cui lun ng nn bi ton c chng minh xong.
Ngoi ra, ta cn c th chng minh bt ng thc trn bng cch sau: S dng bt ng thc
AM GM b ba s, ta c

3xyz
9xyz
2xyz + 1 3 3 x2 y 2 z 2 =

.
3 xyz
x+y+z
Do , ta cn chng minh
x2 + y 2 + z 2 +

9xyz
2(xy + yz + zx).
x+y+z

y chnh l bt ng thc Schur nn bi ton c chng minh xong.


ng thc xy ra khi v ch khi x = y = z = 1. 
Bi 10. Cho a, b, c l cc s thc dng. Chng minh rng
(a + b c)2
(a + c b)2
(c + b a)2
3
+
+
.
2
2
2
2
2
2
(a + b) + c
(a + c) + b
(c + b) + a
5
Chn HSG Quc gia, k Lk, 2014 - 2015
Li gii
(a + b c)2
2c (a + b)
= 1
nn bt ng thc cn chng minh tng ng
2
(a + b) + c2
(a + b)2 + c2
vi bt ng thc Bi 8.
Bi ton c chng minh xong. 

rng

Bi 11. Chng minh bt ng thc sau


3(x2 x + 1)(y 2 y + 1) 2(x2 y 2 xy + 1), x, y R.
Du "=" xy ra khi no?
Chn HSG Quc gia, Qung Tr, 2014 - 2015
Li gii
Do vai tr ca x, y l nh nhau, nn d on ng thc xy ra khi x = y. Khi ta c
3(x2 x + 1)2 = 2(x4 x2 + 1), tng ng vi
(x2 3x + 1)2 = 0.

3 5
T x = y =
. Quay tr li bi ton, bt ng thc cn chng minh tng ng
2
x2 y 2 xy + 1 + 3(x + y)2 3xy(x + y) 3(x + y) 0,
hay
P 2 P + 1 + 3S 2 3SP 3S 0.
11

Nu coi y l mt bt ng thc bc hai theo P th ta c


P = (1 + 3S)2 4(3S 2 3S + 1)
= 3S 2 + 18S 3.
Nu S 0 th P < 0 nn f (P ) > 0. Do ta ch cn xt trng hp S > 0.
Trong trng hp S > 0, li coi bt ng thc trn l mt bt ng thc bc hai theo S, khi ta
c
S = 9(1 + P )2 12(P 2 P + 1)
= 3P 2 + 30P 3.
Nu P 0th S < 0 nn f (S) > 0. Do ta ch cn xt P > 0 l .
Nu P 4 P + 1 > 0 tc l (P + 1)2 > 16P hay P 2 10P + 1 > 4P , khi
(
)
S = 3 P 2 10P + 1 < 12P < 0,
nn suy ra
f (S) > 0.

Nu P 4 P + 1 0, v ch S 2 4P nn S 2 P th ta c
f (S) = 6S 3P 3

12 P 3P 3

= 3(P 4 P + 1)

0, S > 2 P .

Do f (S) l mt hm ng bin trn [2 P ; +), nn f (S) f (2 P ), tc l

f (S) P 2 P + 1 + 3 4P 6P P 6 P

2
= (P 3 P + 1)
0.
Bt ng thc cui lun ng nn bi ton chng
minh xong.
3 5
ng thc xy ra khi v ch khi x = y =
.
2
Bi 12. Cho x, y, z l cc s thc khng m v i mt phn bit. Chng minh rng
x+y
y+z
z+x
9
.
2 +
2 +
2
x+y+z
(x y)
(y z)
(z x)
Chn HSG Quc gia, Chuyn H S phm H Ni, 2014 - 2015
Li gii
Khng mt tnh tng qut, gi s x y z 0. Khi ta c
1 z(3y z)
1
y+z
+
2 =
2 ,
y y(y z)
y
(y z)
z+x
1 z (3x z)
1
+
.
2 =
2
x
x
(z x)
x(z x)

12

Kt hp nh gi trn v s dng bt ng thc AM GM , ta c


(x + y + z)(x + y)
V T (x + y + z)
+ (x + y + z)
(x y)2
(x + y)2 (x + y)2

+
xy
(x y)2

1 1
+
x y

4xy
(x y)2
+
4
+
xy
(x y)2

(x y)2
4xy
5+2

xy
(x y)2

=1+

= 9.
Php chng minh hon tt.

Vi x y z 0 th ng thc xy ra khi v ch khi x = (2 + 3)y, z = 0. 


Bi 13. Cho a, b, c l cc s thc dng tha mn a + b + c = 3. Chng minh rng
)
(
1 1 1
3
3
3
3(ab + bc + ca).
a +b +c +2
+ +
a b c
Chn HSG quc gia, Lm ng, 2014 - 2015
Li gii
S dng bt ng thc Cauchy Schwarz ta c
1 1 1
9
+ +
= 3.
a b c
a+b+c
Do ta cn chng minh
a3 + b3 + c3 + 6 3(ab + bc + ca).
S dng bt ng thc AM GM d thy rng a3 + 1 + 1 3a, do ta cn chng minh
3(a + b + c) 3(ab + bc + ca),
hay tng ng
(a + b + c)2 3(ab + bc + ca).
Hin nhin ng. Bi ton c chng minh xong.
ng thc xy ra khi v ch khi a = b = c = 1. 
Bi 14. Cho ba s khng m a, b, c. Chng minh rng:

5a2 + 4bc + 5b2 + 4ca + 5c2 + 4ab 3(a2 + b2 + c2 ) + 2( ab + bc + ca).


Chn HSG quc gia, Qung Nam, 2014 - 2015
Li gii
Bt ng thc cn chng minh tng ng
(
)
5a2 + 4bc 2 bc 3 (a2 + b2 + c2 ),
13

5a2
(

) 1.
3 (a2 + b2 + c2 )
5a2 + 4bc + 2 bc

S dng bt ng thc AM GM , ta c

8a2 + 3b2 + 3c2 + 4bc


+ +

+ 4bc
,
2

3(a2 + b2 + c2 ) 2 bc a2 + b2 + c2 + 3bc.
3(a2

b2

c2 )

5a2

Do

3 (a2 + b2 + c2 )

(
) 8a2 + 3b2 + 3c2 + 4bc
5a2 + 4bc + 2 bc
+ a2 + b2 + c2 + 3bc
2
10a2 + 5(b + c)2
=
.
2

T , s dng nh gi trn v kt hp vi bt ng thc quen thuc (b + c)2 2(b2 + c2 ), ta c

5a2
10a2
(
)

10a2 + 5(b + c)2


3 (a2 + b2 + c2 )
5a2 + 4bc + 2 bc

2a2
2a2 + (b + c)2

2a2

2a2 + 2(b2 + c2 )
b2
c2
a2
+
+
= 2
a + b2 + c2 a2 + b2 + c2 a2 + b2 + c2
= 1.
=

Bi ton c chng minh xong.


ng thc xy ra khi v ch khi a = b = c = 1. 

Bi 15. Cho ba s thc dng x, y, z tha mn 2 xy + xz = 1. Chng minh rng:


3yz 4zx 5xy
+
+
4.
x
y
z
Chn HSG quc gia, Tuyn Quang, 2014 - 2015
Li gii
Nhn bt ng thc khng c dng i xng, nn ban u mnh on ng thc xy ra khi cc bin
khng bng nhau. Sau mt hi suy ngh khng tm c ng thc xy ra khi no, mnh th
1
cho trng hp x = y = z v u mai gt, n xy ra khi x = y = z = .
3
S dng bt ng thc AM GM , ta c

3x + 2y + z
x+z

=
,
1 = 2 xy + xz x + y +
2
2
t suy ra 3x + 2y + z 2. Nh vy, ta s tm cch nh gi sao cho
3yz 4zx 5xy
+
+
2 (3x + 2y + z).
x
y
z
14

yz zx
xy zx
xy yz

= z,

= x,

= y nn nh gi v tri v x, y, z th ta s s dng
x y
z y
z x
bt ng thc AM GM cho hai s. S dng bt ng thc AM GM , ta c
(
)

yz
zx
xy
(a + c)
+ (b + e)
+ (d + f )
2
ef x + cd y + ab z .
x
y
z

Ch rng

V ta d on c du bng xy ra khi x = y = z nn du bng ca bt ng thc AM GM


tha mn, ta cn c a = b, c = d, e = f . Mt khc theo gi thit, ta phi c a + c = 3, b + e = 4,
d + f = 5. T suy ra e = f = 3, c = d = 2, a = b = 1. Nh vy, ta trnh by nh sau
)
(
)
(
( yz xy )
yz zx
xz xy
+
+2
+
+3
+
VT =
x
y
x
z
y
z
2z + 4y + 6x
= 2 (3x + 2y + z)
4.
1
Bi ton c chng minh xong. ng thc xy ra khi v ch khi x = y = z = . 
3
Bi 16. Cho x, y, z l cc s thc dng tha mn x + y + z = xyz. Chng minh rng
2
1
1
9

+
+
.
4
1 + x2
1 + z2
1 + y2
Chn HSG quc gia, Thi Nguyn, 2014 - 2015
Li gii
Cch 1:
1
1
1
t a = , b = , c = ta c ab + bc + ca = 1. Khi
x
y
z
2a
b
c

+
+
1 + a2
1 + b2
1 + c2
2a
b
c
=
+
+
.
(a + b)(a + c)
(b + a)(b + c)
(c + a)(c + b)
9
. Nhn du cn nh th lm ta nh n
4
ngay bt ng thc AM GM . Tuy nhin ta khng th s dng bt ng thc AM GM kiu

a+b+a+c
bi v khi s cho ra mt nh gi , m ta cn y l .
nh (a + b)(a + c)
2
Ch l

1
1
1

,
=
a+b a+c
(a + b)(a + c)

n y, ta cn nh gi sao cho n b hn hoc bng

nn ta s nh gi bng AM GM kiu nh

(
)
1
1
1
1
1

+
.
a+b a+c
2 a+b a+c
Tuy nhin ta khng th nh gi ba c. V ta cha bit du ng thc xy ra khi no. n
lc d on ng thc t c khi no.
V bt ng thc cn chng minh i xng vi hai bin b v c, nn ta d on ng thc t c
khi b = c. Khi th
1
1
1
1
1
=
,
=k
=k
.
a+b
a+c b+c
b+a
a+c
15

Ta s tm k. S dng bt ng thc AM GM , ta c
b
c
+
+
(a + b)(a + c)
(b + a)(b + c)
(c + a)(c + b)

1
1
b
1
1
c
1
1
=a2

+
k
+
k
a+b a+c
c+a c+b
k (b + a b + c ) k
(
)
)
(
1
1
1
1
b
1
c
1
a
+
+
+k
+
+k
a+b a+c
b+a
b+c
c+a
c+b
2 k
2 k

b
c
b k c k
a+
a+
+
2 k +
2 k + 2
2
=
a+b
a+c
c
+
b

1
2 ka + b
1
2 ka + c
k
=
+
+
.
a+b
a+c
2
2 k
2 k

2 k
1
1
1
biu thc cui cng l mt s khng i th iu kin cn l
= . Suy ra k = . Vi k =
1
1
4
4
th

1
2 ka + b
1
2 ka + c
k
9

+
+
= .
a+b
a+c
2
4
2 k
2 k
9
Con s chnh l iu chng ta mong mun. Bi ton c chng minh xong.
4
7
1
ng thc xy ra khi v ch khi ab + bc + ca = 1, a = 7b = 7c tng ng a = , b = c =
15
15

15
hay x =
, y = z = 15. 
7
Cch 2:
Vi iu kin ab + bc + ca = 1, ta nh n cng thc lng gic trong tam gic

2a

tan A tan B + tan B tan C + tan C tan A = 1.


Do , ta c th t a = tan A, b = tan B, c = tan C, vi A, B, C l ba gc ca mt tam gic. Khi
, ta cn chng minh
9
2 cos A + cos B + cos C .
4
(
)
A
B+C
A
Tht vy, bng mt vi php bin i lng gic, vi ch cos
= cos

= sin , ta c
2
2
2
2
2 cos A + cos B + cos C
(
)
BC
B+C
2A
+ 2 cos
cos
= 2 1 2sin
2
2
2
(
)
A
BC
A
= 2 2sin2 cos
sin
+2
2
2
2
[(
(
)2
)2 ]

A
A
1
BC
1
BC
1
BC
cos
= 2
2 sin
2 2 sin cos
+
+2
+ cos2
2
2 2 2
2
2
4
2
2 2
(
)2
(
)

A
1
1
BC
2B C
2 sin cos
+
= 2
1 sin
+2
2
2
4
2
2 2
1
+2
4
9
= .
4
Bi ton c chng minh xong. 
16

Bi 17. Cho cc s thc khng m x, y, z tha mn: x2 + y 2 + z 2 = 2. Tm gi tr ln nht ca


M=

x2
y+z
1
+
+
.
x2 + yz + x + 1 z + y + x + 1 xyz + 3
Chn HSG Quc gia, Chuyn Hng Vng - Ph Th, 2014 - 2015
Li gii

u tin, ta s chng minh


x2
x

.
2
x + yz + x + 1
z+y+x+1
Tht vy, v x 0 nn ta ch cn chng minh
x(z + y + x + 1) x2 + yz + x + 1,
xz + xy yz + 1,
2xz + 2xy 2yz + 2,
( 2
)
2xz + 2xy 2yz x + y 2 + z 2 0,
(x y z)2 0, lun ng.
T , ta c
x+y+z
1
+
x + y + z + 1 xyz + 3
1
1
=1
+
x + y + z + 1 xyz + 3
xyz + 2 (x + y + z)
=1
.
(xyz + 3) (x + y + z + 1)

Ta s chng minh
xyz + 2 x + y + z (1)

Cch 1:

2
. t S = x+y.
3
T gi thit ta c S 2 + z 2 = 2 + 2xy nn suy ra 2xy = S 2 + z 2 2. Bt ng thc cn chng minh
tng ng
2xyz + 4 2(x + y + z),
(S 2 + z 2 ) z + 4 2S + 2z,
f (S) = zS 2 2S + z 3 2z + 4 0.
Khng mt tnh tng qut, gi s z l s ln nht trong 3 s x, y, z. D thy z

V z > 0, mt khc

(
)
S = 1 z z 3 4z + 4
(
)
= (z 1)2 z 2 + 2z 1

2
.
0, z
3

Nn t suy ra f (S) 0. Nh vy (1) c chng minh. T suy ra


M 1.
ng thc xy ra khi v ch khi x = 0, y = z = 1 nn gi tr ln nht ca M l 1. 

17

Cch 2:
S dng bt ng thc AM GM , ta c 2 = x2 + y 2 + z 2 y 2 + z 2 2yz nn suy ra yz 1.
S dng bt ng thc Cauchy Schwarz v iu thu c bn trn, ta c
(x + y + z xyz)2 = [x(1 yz) + y + z]2
[
] [
]
x2 + (y + z)2 (1 yz)2 + 1
(
)
= (2 + 2yz) y 2 z 2 2yz + 2
= 4 + 2y 2 z 2 (yz 1)
4.
Nh vy (1) c chng minh. T suy ra
M 1.
ng thc xy ra khi v ch khi x = 0, y = z = 1 nn gi tr ln nht ca M l 1. 
Bi 18. Cho a, b, c l cc s thc dng tha mn a + b + c = 6. Chng minh rng khi ta c
a2 + bc b2 + ca c2 + ab
+
+
a2 + b2 + c2 .
b
c
a
Chn HSG tnh, Hi Phng, 2014 - 2015
Li gii
Khng mt tnh tng qut, gi s b l s nm gia hai s a v c. Bt ng thc cn chng
minh tng ng vi
(a b)2 (b c)2 (c a)2
6(a2 + b2 + c2 )
+
+

2(a + b + c).
b
c
a
a+b+c
S dng bt ng thc Cauchy Schwarz ta c
(a b)2 (b c)2 (c a)2
(a b + b c + a c)2
4(a c)2
+
+

=
.
b
c
a
a+b+c
a+b+c
Do , ta cn phi chng minh
6(a2 + b2 + c2 )
4(a c)2

2(a + b + c),
a+b+c
a+b+c
hay tng ng
4(a c)2 6(a2 + b2 + c2 ) 2(a + b + c)2 ,
4(a c)2 2(a b)2 + 2(b c)2 + 2(c a)2 ,
(a c)2 (a b)2 + (b c)2 ,
2(a b)(b c) 0.
Bt ng thc cui hin hin ng do b l s nm gia hai s a v c. Bi ton c chng minh
xong.
ng thc xy ra khi v ch khi a = b = c = 2. 

18

Bi 19. Cho a, b l 2 s tha mn iu kin: a2 + b2 + 9 = 6a + 2b. Chng minh


4b 3a.
Chn HSG tnh Bnh Thun, 2014 - 2015
Li gii
D on du bng khi 4b = 3a, kt hp vi gi thit a2 + b2 + 9 = 6a + 2b d thy a =

12
9
,b = .
5
5

T d on ta c li gii nh sau:
S dng bt ng thc AM GM , ta c
(

)2
12
24a
a +

,
5
5
( )2
9
18b
2
b +

.
5
5
2

Cng v vi v hai bt ng thc trn, ta thu c


24a + 18b
,
5

a2 + b2 + 9
hay tng ng
6a + 2b

24a + 18b
.
5

T ta c
4b 3a.
Php chng minh hon tt.
ng thc xy ra khi v ch khi a =

12
9
,b = . 
5
5

Bi 20. Cho ba s dng a, b, c tha mn a + b + c = 3. Tm gi tr nh nht ca biu thc


P = 7(a4 + b4 + c4 ) +

ab + bc + ca
.
+ b2 c + c2 a

a2 b

Chn HSG tnh B Ra Vng Tu, 2014 - 2015


Li gii
Khng mt tnh tng qut, gi s b l s nm gia a v c, khi ta c c(a b)(b c) 0,
tng ng
a2 b + b2 c + c2 a b(a2 + ca + c2 ).
T , kt hp vi bt ng thc AM GM b ba s, ta c
( 2
)
a b + b2 c + c2 a (ab + bc + ca) b(a2 + ca + c2 ) (ab + bc + ca)
(3b + a2 + ca + c2 + ab + bc + ca)

34
(
)3
2
(a + c) + 3b + ab + bc
=
34
(
)3
(3 b)2 + 3b + b(3 b)
=
34
= 9.
19

Mt khc, theo bt ng thc Cauchy Schwarz th


a 4 + b 4 + c4

(a2 + b2 + c2 )2
.
3

Do , s dng cc nh gi trn, sau lin tc dng Cauchy Schwarz ta c


)2 (ab + bc + ca)2
7( 2
a + b 2 + c2 +
3
9
2
)2 (a + b2 + c2 )2 + (ab + bc + ca)2 + (ab + bc + ca)2
41 ( 2
=
a + b 2 + c2 +
18
18
2
4
2
2
2
41 (a + b + c)
(a + b + c + ab + bc + ca + ab + bc + ca)

+
18
32
18 3
22
= (a + b + c)4
81
= 22.

ng thc xy ra khi v ch khi a = b = c = 1 nn gi tr nh nht ca P l 22. 


Bi 21. Cho a, b v c l cc s thc dng. Tm gi tr nh nht ca biu thc:
P =

a + 3c
4b
8c
+

.
a + 2b + c a + b + 2c a + b + 3c
Chn HSG tnh Kin Giang, 2014 - 2015
Li gii

a = x + 5y 3z
x = a + 2b + c
. Do , ta cn tm gi tr nh nht ca
t y = a + b + 2c ta c b = x 2y + z

z = a + b + 3c
c = y + z
P =

4x 2y 8y 4z
+
+
+
17.
y
x
z
y

S dng bt ng thc AM GM , ta c

4x 2y
8y 4z
P 2

+2

17
y x
z y

= 12 2 17.

ng thc xy ra khi v ch khi b= (1 + 2)a, c = (4 + 3 2)a.


Vy gi tr nh nht ca P l 12 2 17.
Bi 22. Cho a, b, c l ba s thc dng tha mn abc = 1. Chng minh rng:
a3

1
1
1
+ 3
+ 3
1.
3
3
+ b + 1 b + c + 1 c + a3 + 1
Chn HSG tnh Long An, 2014 - 2015
Li gii

Ta c
a3 + b3 + 1 = (a + b)(a b)2 + ab(a + b) + abc ab(a + b) + abc = ab(a + b + c).
20

Do , ta c
a3

1
1
c

=
.
3
+b +1
ab(a + b + c)
a+b+c

Tng t vi hai biu thc cn li, sau cng v vi v ta thu ngay c iu phi chng minh.
ng thc xy ra khi v ch khi a = b = c = 1. 
Bi 23. Cho cc s thc a, b, c 1 tha mn a + b + c = 6. Chng minh rng:
(a2 + 2)(b2 + 2)(c2 + 2) 216.
Chn HSG tnh Vnh Phc, 2014 - 2015
Li gii
Cch 1:
Khng mt tnh tng qut, gi s a b c, khi d thy a 2 v c 2.
Ta s chng minh rng
(
)2
2
(a + b)
(a2 + 2)(b2 + 2)
+2 .
4
Tht vy, v a2 + 6ab + b2 16 22 + 6.2.1 + 12 16 = 1 > 0 nn
(
)2
2
(
)
(a
+
b)
1
(a2 + 2)(b2 + 2)
+ 2 = (a b)2 a2 + 6ab + b2 16 0.
4
16
Do ta quy bi ton v chng minh
(
)2
(6 c)2
+ 2 (c2 + 2) 216.
4
Tht vy, v 1 c 2 nn
c4 20c3 + 150c2 424c + 104 2c3 20c3 + 300c 424c + 104
= 18c3 124c + 104
18c3 124 + 104
= 18c3 20
< 0.
T ta c
(

(6 c)
+2
4
2

)2
(c2 + 2) = 216 +

(
)
1
(c 2)2 c4 20c3 + 150c2 424c + 104
16

216.
Bi ton c chng minh xong.
ng thc xy ra khi v ch khi a = b = c = 2. 
Chc hn nhiu bn thc mc: Sao phn tch g m khng th?
Thc ra l mnh dng lnh factor trong Maple.
Vy nu trong phng thi th lm th no? Mnh trnh by nh sau:
Xt hm f (c) trn [1; 2], trong
)2
(
2
(6 c)
+ 2 (c2 + 2).
f (c) =
4
21

Ta c

)
(
)2
2
2
(6

c)
(6

c)
(6

c)
f (c) = 2
+2
(c2 + 2) +
+ 2 2c
4
2
4
(
)2 (
)
(6 c)2
(6 c)(c2 + 2)
=
+ 2 2c
(6c)2
4
+2
4
(
)2 ( [
)
]
2
2
2
c
(6

c)
+
8

2(6

c)(c
+
2)
(6 c)
=2
+2
.
4
(6 c)2 + 8

Vi php phn
tch nh
[
] trn th chng ta gim c lng tnh ton rt nhiu v ch cn xt
du ca c (6 c)2 + 8 2(6 c)(c2 + 2) trn (1; 2). Ta c
[
]
c (6 c)2 + 8 2(6 c)(c2 + 2) = 3(c3 8c2 + 16c 8)
= 3(c 2)(c2 6c + 4)

= 3(c 2)(c 3 5)(c 3 + 5)


> 0, c (1, 2).
Do f (c) > 0, c (1, 2) nn hm f (c) ng bin trn [1; 2], t suy ra
f (c) f (2) = 216.
Bi ton c chng minh xong.
ng thc xy ra khi v ch khi a = b = c = 2. 
Cch 2:
t M = (a2 + 2)(b2 + 2)(c2 + 2) ta c
ln M = ln(a2 + 2) + ln(b2 + 2) + ln(c2 + 2).
V a, b, c 1 v a + b + c = 6 nn 1 a, b, c 4. Dng k thut h s bt nh, ta cn chng minh
vi mi t [1; 4] th
f (t) = ln(t2 + 2) (xt + y) 0.
2
iu kin cn bt ng thc ny ng l f (2) = 0 v f (2) = 0, t gii ra c x = v
3
4
y = ln 6 .
3
2
4
iu kin : Xt hm s f (t) = ln(t2 + 2) t ln 6 + trn [1; 4].
3
3
2(t 1)(t 2)

, f (t) = 0 t = 1 hoc t = 2.
Ta c f (t) =
3(t2 + 2)
M f (t) lin tc trn [1; 4] v f (2) = max{f (1), f (2), f (4)} nn f (x) f (2) = 0.
T , ta c
12
2
ln M = f (a) + f (b) + f (c) + (a + b + c) + ln 216
3
3
2
12
(a + b + c) + ln 216
3
3
= ln 216.
Suy ra M 216. Bi ton c chng minh xong.
ng thc xy ra khi v ch khi a = b = c = 2. 

22

Bi 24. Cho a, b, c l cc s thc dng. Tm gi tr ln nht ca biu thc

8a + 3b + 4( ab + bc + 3 abc)
P =
.
1 + (a + b + c)2
Chn HSG tnh, Thanh Ha, 2014 - 2015
Li gii
Nhn xt rng: theo AM GM cho mu s (MS) th ta c M S 2(a + b + c), vy nu t s
k
(TS) ta nh gi c T S k (a + b + c) (k l mt hng s), th khi P v kh nng cao
2
k
chnh l gi tr ln nht ca P . Ti g khng th nh!
2
Nhn TS c cc biu thc cha cn, m ta cn nh gi n b thua hoc bng k (a + b + c) nn ta
s ngh ngay n bt ng thc AM GM . Tuy nhin, ta cha d on c du bng khi no,
nn chng ta s gi s ng thc t c khi a = mb = nc, ta phi tm m v n.
S dng bt ng thc AM GM , ta c

1
a + bm
ab =
a bm ,
m
2 m

1
bm + cn
bc =
bm cn
,
mn
2 mn

1
a + bm + cn
3
3

abc =
a bm cn
.
3
mn
3 3 mn
T ta c

(
)
(
)
(
)

2
4
m
4m
n
4n
TS 8 + +
a+ 3+2 m+2
+
b+ 2
+
c.
n
m 3 3 mn
m 3 3 mn
3 3 mn
T S k (a + b + c) th ta phi c

4
2
=
3
+
2
m+2
k =8+ +
m 3 3 mn

4m
m
+
=2
n
3 3 mn

4n
n
+
. (1)
3
m 3 mn

Ngi trong phng thi m gii c ci h ny tm c m, n th... chi i chc ti cht... Ch


rng y l mt bi trong thi, ngi ra s ra sao cho s c ngi lm c, nn kiu g
h s m, n cng
l s
p ch n khng l tot c. Do m, n l s phi sao cho my ci cn

m
n
m, 3 mn,
,
s tnh ra s p. Ta s ch n thng m hn v n c mt trong c 4
n
m

ci cn, trong 4 ci cn c ci m nn ta ch xt m = 1; 4; 9; 16; 25; ...


Nu m = 1 th

4m
4n
m
n
+
+
3+2 m+2
>2
, khng tha mn iu kin.
3
3
n
m 3 mn
3 mn

Nu m = 4, mun ci 3 mn p th n = 2; 16. Nhng vi n = 2 th ci n khng p, nn n = 16.


Thay m = 4, n = 16 vo thy n hon ton tha mn (1). Tht may mn!!!
Tuy rng suy lun khng hon ton thuyt phc, nhng cng thm cht may mn kt qu li c
nh . Trong vic g cng vy, dm ngh, dm lm, thm cht may mn th thnh cng.
28
Quay tr li bi ton, vi m = 4, n = 16 th thay vo (1) ta c k = . Nh vy, s dng nh
3
gi on u ta s c
k
14
P = .
2
3
23

16
4
1
ng thc xy ra khi v ch khi a + b + c = 1, a = 4b = 16c hay a = , b = , c = .
21
21
21
14
Vy, gi tr ln nht ca P l
.
3
Bi 25. Cho x, y, z l cc s thc dng tha mn xy + yz + zx = 2xyz. Chng minh rng:

x
y
z
+
+
1.
2
2
2
2
2
2
2y z + xyz
2z x + xyz
2x y + xyz
Chn HSG tnh, Gia Lai, 2014 - 2015
Li gii
1
1
1
,b = ,c =
th ta c a, b, c > 0 v a + b + c = 2. Bt ng thc cn chng
x
y
z
minh tng ng
bc
ca
ab

+
+
1.
2a + bc
2b + ca
2c + ab

t a =

S dng bt ng thc AM GM , ta c

Tng t, ta thu c

bc
bc
=
2a + bc
(a + b + c)a + bc
bc
=
(a + b)(a + c)
(
)
1
bc
1

+
.
2 a+b a+c

(
)
1
ca
ca
1

+
,
2 b+c b+a
2b + ca
(
)
ab
1
ab
1

+
.
2 c+a c+b
2c + ab

Cng v vi v cc bt ng thc trn, ta thu c


(
)
(
)
(
)
ab
1
1
bc
1
1
ca
1
1
VT
+
+
+
+
+
2 c+a c+b
2 a+b a+c
2 b+c b+a
(
)
1 ab + bc ab + ca bc + ca
=
+
+
2
a+c
c+b
a+b
a+b+c
=
2
= 1.
Bi ton c chng minh xong.
ng thc xy ra khi v ch khi a = b = c =

3
2
hay x = y = z = . 
3
2

24

Bi 26. Cho x, y, z l cc s thc dng tha mn iu kin x 1, y 2 v x + y + z = 6. Chng


minh rng
(x + 1) (y + 1) (z + 1) 4xyz.
thi chuyn h lp 10, THPT Chuyn S phm, 2014 - 2015
Li gii
Bt ng thc cn chng minh tng ng vi
x + y + z + xy + yz + zx + 1 3xyz,
hay
7 + z(6 z) + xy(1 3z) 0.
V x 1, y 2 nn z 3, tc l 1 3z < 0 v 3z 5 > 0.
S dng bt ng thc AM GM , ta c
xy =

1
(2x + y)2
(1 + x + y)2
(7 z)2
2x y

=
.
2
8
8
8

Do
7 + z(6 z) + xy(1 3z) 7 + z(6 z) +

(7 z)2
(1 3z)
8

1
= (z 3)(7 z)(3z 5)
8
1
= (z 3)(1 + x + y)(3z 5)
8
0.
Bi ton c chng minh xong.
ng thc xy ra khi v ch khi x = 1, y = 2, z = 3. 
Bi 27. Cho cc s thc dng a, b, c tha mn a + b + c = 1. Chng minh rng
(
)
1+a 1+b 1+c
a b c
+
+
2
+ +
.
1a 1b 1c
b c a
Chn i tuyn Olympic Ton lp 10 vng 1, Chuyn Nguyn Du, 2014 - 2015
Li gii
Cch 1:
S dng bt ng thc Cauchy Schwarz ta c
a b c
(a + b + c)2
+ +
.
b c a
ab + bc + ca
Mt khc, ta c
1+a 1+b 1+c
a+b+c+a a+b+c+b a+b+c+c
+
+
=
+
+
1a 1b 1c
a + c)
a+b
( b+c
b
c
a
+
+
=2
+ 3.
b+c c+a a+b
Do , ta quy bi ton v chng minh bt ng thc mnh hn l
a
b
c
3
(a + b + c)2

+
+
+ .
ab + bc + ca
b+c c+a a+b 2
25

rng
(a + b + c)2
(a b)2 + (b c)2 + (c a)2
3=
,
ab + bc + ca
2(ab + bc + ca)
b
c
3
(a b)2
(b c)2
(c a)2
a
+
+
+ 3=
+
+
.
b+c c+a a+b 2
2(a + c)(b + c) 2(a + b)(a + c) 2(b + c)(b + a)
Do ta cn phi chng minh
Sc (a b)2 + Sa (b c)2 + Sb (c a)2 0,
trong

1
a2
1

=
0
S
=

2 (ab + bc + ca) 2(a + b)(a + c)


2 (ab + bc + ca) (b + c)(a + b)

1
1
b2
Sb =

=
0

2 (ab + bc + ca) 2(b + c)(b + a)


2 (ab + bc + ca) (b + c)(b + a)

1
1
c2

Sc =

=
0
2 (ab + bc + ca) 2(c + a)(c + b)
2 (ab + bc + ca) (c + a)(c + b)
Vy bt ng thc cui lun ng, bi ton c chng minh xong.
1
ng thc xy ra khi v ch khi a = b = c = . 
3
Cch 2:
Theo cch 1, ta quy bi ton v chng minh bt ng thc mnh hn l
(a + b + c)2
a
b
c
3

+
+
+ .
ab + bc + ca
b+c c+a a+b 2
Nhn hai v vi ab + bc + ca, ta cn chng minh
(
)
a
b
c
3
2
(a + b + c) (ab + bc + ca)
+
+
+ (ab + bc + ca) .
b+c c+a a+b
2
(
)
1
1 1 1
S dng bt ng thc quen thuc

+
, ta c
x+y
4 x y
(
)
(
)
a
b
c
1
1
1
2
2
2
+
+
= a + b + c + abc
+
+
(ab + bc + ca)
b+c c+a a+b
b+c c+a a+b
(
)
abc 1 1 1 1 1 1
2
2
2
a +b +c +
+ + + + +
4
b c c a a b
ab + bc + ca
.
= a2 + b2 + c2 +
2
T suy ra
V P a2 + b2 + c2 +

ab + bc + ca 3
+ (ab + bc + ca)
2
2

= (a + b + c)2 .
chnh l iu cn chng minh.
1
ng thc xy ra khi v ch khi a = b = c = . 
3

26

Bi 28. Cho a, b, c, d l cc s thc dng tha mn a + b + c + d = 4. Chng minh rng

(a + b)2
(b + c)2
(c + d)2
(d + a)2
P =
+
+
+
16.
a2 ab + b2
b2 bc + c2
c2 cd + d2
d2 ad + a2
thi kho st i tuyn lp 10 vng 2, Chuyn KHTN, 2014 - 2015
Li gii
Ta c a2 ab + b2 =

(a + b)2 3(a b)2


(a + b)2
+

v (a + b)2 (a + b)(a + 1) nn
4
4
4
2
(a + b)

2(a + 1).
a2 ab + b2

Thit lp ba biu thc cn li, sau cng v vi v v ch a + b + c + d = 4 ta thu ngay c


iu phi chng minh.
ng thc xy ra khi v ch khi a = b = c = d = 1. 
Bi 29. Cho x, y, z l cc s thc dng. Chng minh:
(
)
x (
y)(
z)
x+y+z
1+
1+
1+
2+2
.
3 xyz
y
z
x
Chn i tuyn d thi Olympic 30-4 lp 10, tnh Bnh Thun, 2014 - 2015
Li gii
Bt ng thc cn chng minh tng ng vi
(
) (
)
x y z
x y z
x+y+z
+ +
+
+ +
2
.
3 xyz
y z x
z x y
S dng bt ng thc AM GM , ta c

x z
z
z2
3z
+ + 33
=
,
3 xyz
y x x
xy

y x x
x2
3x
+ + 33
=
,
3 xyz
z y y
yz

2
3y
z y y
3 y
+ + 3
=
.
3
x z z
xz
xyz

Cng v vi v ba bt ng thc trn, ta thu c


x y z
x+y+z
+ +
.
3 xyz
y z x
Hon ton tng t, ta cng chng minh c
x y z
x+y+z
+ +
.
3 xyz
z x y
T cng v vi v hai bt ng thc trn ta thu c iu phi chng minh.
ng thc xy ra khi v ch khi x = y = z. 

27

Trong qu trnh lm khng trnh khi sai st, rt mong nhn c phn hi t bn c gn xa
ti liu c hon thin hn. Mi kin ng gp xin gi v hm th: liltee.spvl@gmail.com. Xin trn
trng cm n.
Tng Hi Tun
Lil.Tee
http://tanghaituan.com
http://ask.fm/TangHaiTuanVLPT
https://facebook.com/tanghaituan.vlpt

28

Ti liu
[1] thi c ly ti chuyn mc Thi HSG cp Tnh, Thnh ph. Olympic 30-4. thi v kim
tra i tuyn cc cp ca Din n ton hc http://diendantoanhoc.net.
[2] V Quc B Cn, Li gii v bnh lun thi Olympic qua cc nm, Bi 51 trang 20.
[3] http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3712545#p3712545

29

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