100 bất đẳng thức ôn thi đại hoc

100 INEQUALITY PROBLEMS
*****
Cao Minh Quang
05/11/2006

Most of these problems were collected from the Mathematics and Youth Magazine in Vietnam.
Translation from Vietnamese into English may have many errors. I am looking forward to hearing
from your ideas.
• Address: Cao Minh Quang, Mathematics teacher, Nguyen Binh Khiem specialized High School,
Vinh Long town, Vinh Long, Vietnam.
• Email:

1.
()
a,b 0,a b 1>+=,
22
ab4
a1b15
+
≤
++
.
First solution.
Applying the AM – GM Inequality we get
22 2
2
1 3 1 3 4a 3 a 4a
a1a 2a.
44 44 4 a 14a3
+
+= + + ≥ + = ⇒ ≤
+
+

.
Similarly,
2
b
4b
4b 3
b1
≤
+
+
.
Adding these two inequalities,
22
ab 4a4b 1 1
23
a 1b 1 4a34b3 4a34b3
⎡
⎤
⎛⎞⎛⎞
+≤ + =− +
⎜⎟⎜⎟
⎢
⎥
++ ++ ++
⎝⎠⎝⎠
⎣
⎦
.
On the orther hand,
()()

()
11 11 4 2
4a 3 4b 3 4
4a 3 4b 3 4a 3 4b 3 4 a b 6 5
⎡⎤
⎛⎞⎛⎞
++ + + ≥⇒ + ≥ =⎡⎤
⎜⎟⎜⎟
⎢⎥
⎣⎦
++ ++ ++
⎝⎠⎝⎠
⎣⎦
.
Thus,
22
ab 24
23.
a1b1 55
+
≤− =
++
.
Second solution.
Applying the AM – GM Inequality we get
4
22 2 2
5
5
1111 1 5

a1a 5a 4a
4444 4 4
⎛⎞
+= ++++≥ =
⎜⎟
⎝⎠
.
Similarly,
22
5
5
b
14b
4
+≥
.
Adding these two inequalities,
33
55
22
22
55
ab a b411 11
a. . b. .
55
a1b1 5 22 22
4a 4b
44
⎛⎞
+≤ + = + ≤

⎜⎟
⎜⎟
++
⎝⎠

()
11 11
aaa bbb
3a b 2
444
22 22
55 5 555
⎡⎤
⎛⎞⎛⎞
++++ ++++
⎢⎥
⎜⎟⎜⎟
++
⎡⎤
≤+==
⎢⎥
⎜⎟⎜⎟
⎢⎥
⎢⎥
⎣⎦
⎜⎟⎜⎟
⎜⎟⎜⎟
⎢⎥
⎝⎠⎝⎠
⎣⎦

.
Third solution.
Applying the AM – GM Inequality we get
1
ab2ab ab
4
+
≥⇒≤
.
Therefore,
()
22
2
22 2222
22
a b ab a b a b ab 1
a1b1aba b1
ab 2abab 1
+++ +
+= = =
++ +++
+−+ +

2
22
1
1
ab 1 ab 1 4
4
31 31

ab 2ab 2 5
1331
.
ab ab
24 16
4216
+
++
== ≤=
−+
⎛⎞
−+
−−+
⎜⎟
⎝⎠
.
2.
()
a,b,c 0,a b c 1>++=,
abc3
a1b1c1 4
+
+≤
+
++
.
First solution.
Applying the AM – GM Inequality we get
()()
aa1aa

a1 a b ac 4ab ac
⎛⎞
=≤+
⎜⎟
++++ ++
⎝⎠
,
()()
b
b1bb
b
1ba bc4babc
⎛⎞
=≤+
⎜⎟
++++ ++
⎝⎠
,
()()
cc1cc
c1 cb ca 4c b ca
⎛⎞
=≤+
⎜⎟
++++ ++
⎝⎠
.
Adding these three inequalities,
abc1abbcac3
a1b1c1 4ab bc ac 4

+++
⎛⎞
++≤ ++ =
⎜⎟
+++ + ++
⎝⎠
.
Second solution.
Applying the AM – GM Inequality we get
+=+ + + ≥ ⇒ ≤
+
4
33
4
3
111 1 a 1
a1 a 4a a3
333 a14
3
,
Similarly,
4
33
b1
b3
b1 4
≤
+
,
4

33
c1
c3
c1 4
≤
+
.
Adding these three inequalities,
44 4
abc3 1 1 1
a.a.a. b.b.b. c.c.c.
a1 b1 c1 4 3 3 3
⎛⎞
++≤ + + ≤
⎜⎟
⎜⎟
+++
⎝⎠

111
aaa bbb ccc
33
333
44 4 4 4
⎡⎤
⎛⎞⎛⎞⎛⎞
+++ +++ +++
⎢⎥
⎜⎟⎜⎟⎜⎟
≤++=

⎢⎥
⎜⎟⎜⎟⎜⎟
⎢⎥
⎜⎟⎜⎟⎜⎟
⎢⎥
⎝⎠⎝⎠⎝⎠
⎣⎦
.
Third solution.
Applying the AM – GM Inequality we get

()()()
⎛⎞
++ +++ + + ≥
⎡⎤
⎜⎟
⎣⎦
+++
⎝⎠
111
a1 b1 c1 9
a1 b1 c1

()()()
⎛⎞
⇒++≥ =
⎜⎟
+++ +++++
⎝⎠
111 9 9

a1 b1 c1 a1 b1 c1 4
.
Therefore,
abc 111 93
33
a1 b1c1 a1 b1c1 4 4
⎛⎞
++=− ++ ≤−=
⎜⎟
+++ +++
⎝⎠
.
Forth solution
Applying the AM – GM Inequality we get
+=+ + ≥ + ⇒ ≤ + −
+
+
12 a2 a 3a11 1
a1 a 2 .
33 33 a1 2 23
a2
2
33
.
Similarly,
b3b111
.
b1 2 2 3
b2
2

33
≤+−
+
+
,
c3c111
.
c1 2 2 3
c2
2
33
≤+−
+
+
.
Adding these three inequalities,
abc3abc311 1 1
a1 b1 c1 2 3 3 3 2 3
a2 b2 b2
222
33 33 33
⎛⎞
⎜⎟
⎛⎞
⎜⎟
++≤ +++− + + ≤
⎜⎟
⎜⎟
+++
⎜⎟

⎝⎠
+++
⎜⎟
⎝⎠


111
abc
33993
333
3
22 2 2 2 44
abc 2
23.
333 3
⎡⎤
⎛⎞⎛⎞⎛⎞
+++
⎢⎥
⎜⎟⎜⎟⎜⎟
≤+++− ≤−=
⎢⎥
⎜⎟⎜⎟⎜⎟
⎛⎞
⎢⎥
⎜⎟⎜⎟⎜⎟
++ +
⎜⎟
⎢⎥
⎝⎠⎝⎠⎝⎠

⎣⎦
⎝⎠
.
3.
()
a,b,c 1≥ ,
()
111
2abc 9
abc
⎛⎞
+
++ ≥
⎜⎟
⎝⎠

Solution.
We have,
()
(
)
a1b1 0 ab1a b−−≥⇔+≥+,
()
(
)
ab 1 c 1 0 abc 1 ab c−−≥⇔+≥+.
Adding these two inequalities, we obtain
abc 2 a b c
+
≥++.

Thus,
() ()
111 111
2abc abc 9
abc abc
⎛⎞ ⎛⎞
+++≥++++≥
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
.
4.
()
x,y,z0,xyxyyzyzzxzx1>++=,
+
+≥
+++
666
33 33 33
x
y
z1
x
yy
zzx2
.
Solution.
We set
333
a x , b y , c z=== and observe that
ab bc ca 1

+
+=.
The inequality is equivalent to
+
+≥
+++
222
abc1
ab bc ca 2
.
Applying the Cauchy – Buniakowski Inequality we get
()()()( )
⎡⎤
++ +++++≥++
⎡⎤
⎢⎥
⎣⎦
+++
⎣⎦
222
2
abc
ab bc ca abc
ab bc ca

()
()
⇒++≥++≥ ++=
+++
222

abc1 1 1
abc ab bc ca
ab bc ca 2 2 2
.
5.
()
x, y 0,xy 1>=
,
++++ ≥
++
22
22
9
x3xy3y 11
xy1
.
Solution.
Applying the AM – GM Inequality we get
() ()
+++ ++−≥ ++ −=
++ ++
22 22
4
22 22
99
x y 1 3x 3y 1 4 x y 1 . .3x.3y 1 11
xy1 xy1
.
6.
{

}
()
m, n \ 0 , m a, b, c n∈≤≤ ,
()
()
2
nm
abc3
bcacab 22mnm
−
++≤+
+++ +
.
Solution.
Without loss of generality, we can assume that
abc≥≥
, we set
xbc, yca, zab
=
+=+=+
.
We observe that
xyz≤≤, therefore,

⎛⎞⎛⎞
⎛⎞ ⎛⎞
−
−+− −≥
⎜⎟⎜⎟
⎜⎟ ⎜⎟

⎝⎠ ⎝⎠
⎝⎠⎝⎠
yz xy
11 11 0
xy yz

⎛⎞
+++
⇔++≤++
⎜⎟
⎝⎠
yz xz xy x z
22
xyz yx


()
()
−
⎛⎞
⇔+++≤ +
⎜⎟
+++
⎝⎠
2
zx
abc
23261
bc ca ab xy
.

On the other hand,
zx ac nm
0
xbc2m
−−−
≤=≤
+
and
zx ac ac nc nm
0
zabacncnm
−
−−−−
≤=≤≤≤
+
+++
.
Thus,
()()
()
()
−−
≤
+
22
zx nm
2
xz 2m n m
.
From (1) and (2), we have

()
()
2
nm
abc3
bcacab 22mnm
−
++≤+
+++ +
.
7.
()
12 n
0 x , x , , x 1, n 2<≤≥,
()()()
12 n
12 n
xx x
1
n1n1x1x 1x
+++
≤
+− − −
.
Solution.
We set
()
−= =
ii
1 x a , i 1, 2, , n and observe that

i
0a 1
≤
< ,
(
)
=i 1, 2, , n .
The inequality is equivalent to
()()
(
)
12 n
12 n
1a 1a 1a
1
n 1 na a a
−+−++−
≤
+
,
or
()
(
)
2
12 n 12n 12n
a a a 1 a a a n a a a+++ + ≥ .
Applying the AM – GM Inequality we get
n
12 n 12n

a a a n a a a+++≥ and
() ( )
n1
n
12 n 12 n 12 n
1 na a a 1 n 1 a a a n a a a
−
+≥+− ≥
.
Therefore,
()()()()()
n
22 2
n
1 2 n 12n 12n 12 n 12n 12n
a a a 1 a a a n a a a n a a a 1 a a a n a a a+++ + ≥ = +++ + ≥
.
8.
()
0xyz1,3x2yz4<<≤≤ + +≤ ,
+
+≤
222
10
3x 2y z
3
.
Solution.
We have 3x 2y z 4++≤, thus, ++≤
2

3x 2xy xz 4x .
On the other hand,
0xyz1<<≤≤, therefore
(
)
(
)
2
yy
x2
y
x
−
≤−
and
()
zz x z x−≤−
.
Adding these two inequalities, we obtain
+
+≤++
222
3x 2y z x 2y z .
Applying the Cauchy – Buniakowski Inequality we get
()
()
()
⎛⎞
++ ≤++≤++ ++⇒++≤
⎜⎟

⎝⎠
2
2
222 222 222
110
3x 2y z x 2y z 2 1 3x 2y z 3x 2y z
33
.
9.
()
0xyz1≤≤≤≤,
()()()
−+ −+ −≤
222
108
xyz yzy z1z
529
.
Solution.
We set
()()
(
)
=−+−+−
222
Txyz yzy z1z.
Applying the AM – GM Inequality and the Cauchy – Buniakowski Inequality we get
()() ()
3
22

11yy2z2y
T0 y.y2z2y z1z z1z
223
++ −
⎛⎞
≤+ − + − ≤ + −
⎡⎤
⎜⎟
⎣⎦
⎝⎠


2
22
42354232323
zz1zz1z zz1z
27 27 23 54 54 27
⎛ ⎞⎛ ⎞⎛⎞⎛⎞⎛⎞⎛ ⎞
=+−=−= −
⎜ ⎟⎜ ⎟⎜⎟⎜⎟⎜⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝⎠⎝⎠⎝⎠⎝ ⎠


2
23 23 23
zz1z
54 108
54 54 27
23 3 529
⎡⎤

++−
⎢⎥
⎛⎞
≤=
⎢⎥
⎜⎟
⎝⎠
⎢⎥
⎣⎦
.
10.
()
222
a,b,c ,abc8∈++≤ ,
+
+≥−ab bc 2bc 8 .
Solution.
We have
()
2
2
222
b3b
8 ab bc 2bc a b c ab bc 2bc a c 0
24
⎛⎞
+++ ≥+++++ =+++ ≥
⎜⎟
⎝⎠
.

Therefore,
ab bc ca 8++≥−.
11.
()
a,b 0> ,
()()()
()
()
2
22
ab 3aba3b
ab
ab
2
8a b a 6ab b
−++
+
≥+
+++
.
Solution.
If
ab= , the inequality is true.
If
ab≠ , the inequality is equivalent to
()()()
()
()
−++
+

−≥
+++
2
22
ab 3aba3b
ab
ab
2
8a b a 6ab b


()
(
)
()()
()
()
⎡⎤
+++
⎢⎥
⇔− − ≥
⎢⎥
+++
⎢⎥
⎣⎦
2
2
22
ab3aba3b
ab1 0

4a b a 6ab b


()
(
)
(
)
(
)
⇔+ ++−++ ++ ≥
22 22
4 a b a 6ab b a b 2 ab 3a 3b 10ab 0 .
We set
xab, y ab=+ = and observe that
22 2 2
abx2y+=− .
The inequality is equivalent to
()
()
()
()
+−+ +≥⇔−≥⇔≥⇔+≥
3
22 22
4x x 4
y
x2
y
3x 4

y
0x2
y
0x2
y
ab2ab.
12.
()
x ∈  ,
33
sin x sin 2x sin3x
2
++<.
Solution.
We have,
++=+ =+ ≤
2
sin x sin 2x sin3x sin 2x 2sin 2x cos x sin 2x 4sin x cos x

22
2242
cos x cos x
1 4sinxcos x 1 4 sin xcos x 1 8 sin x. .
22
≤+ ≤+ =+


3
22
2

cos x cos x
sin x
83 33
22
18 1
392
⎛⎞
++
⎜⎟
≤+ =+ <
⎜⎟
⎜⎟
⎜⎟
⎝⎠
.
13.
()
12 n
x , x , , x 0,n 3>≥,
()() ()()
22
22
123 234
n1 n 1 n 1 2
12 3 23 4 n1n 1 n1 2
xxx xxx
xxx xxx
n
xx x xx x x x x x x x
−

−
++
++
+++ +≥
++ ++
.
Solution.
We set,
()() ()()
22
22
123 234
n1 n 1 n 1 2
12 3 23 4 n1n 1 n1 2
xxx xxx
xxx xxx
A
xx x xx x x x x xx x
−
−
++
++
=+++ +
++ ++
.
Applying the AM – GM Inequality we have
() () () ()
22
22
123 234

n1 n 1 n 1 2
12 3 23 4 n1n 1 n1 2
xxx xxx
xxx xxx
A n 1 1 1 1
xx x xx x x x x x x x
−
−
⎡⎤⎡⎤⎡ ⎤⎡⎤
++
++
+= ++ +++ ++ +
⎢⎥⎢⎥⎢ ⎥⎢⎥
++ ++
⎣⎦⎣⎦⎣ ⎦⎣⎦

()()
()
()()
()
(
)
(
)
()
()()
()
1213 2324 n1nn11 n1n2
12 3 23 4 n1n 1 n1 2
xxxx xxxx x xx x xxxx

xx x xx x x x x xx x
−−
−
++ ++ + + ++
=+ ++ +
++ + +

()( )( )( )
n
13 24 n11 n2
1324 n11n 2
n
n
12 n1n 12 n1n
2xxxx xxxxx x x x x x x x
nn2n
x x x x x x x x
−
−
−−
++ ++
≥≥=
.
Therefore,
()() ()()
22
22
123 234
n1 n 1 n 1 2

12 3 23 4 n1n 1 n1 2
xxx xxx
xxx xxx
n
xx x xx x x x x xx x
−
−
++
++
+++ +≥
++ ++
.
14.
()
≤≤1a,b,c2,
()
111
abc 10
abc
⎛⎞
+
+++≤
⎜⎟
⎝⎠
.
Solution.
Without loss of generality, we can assume that
1abc2
≤
≤≤≤, we obtain,

ab bc abbcac
11 11 0 2
bc ab bcabca
⎛⎞⎛⎞⎛⎞⎛⎞
− − +− − ≥⇔+++≤++
⎜⎟⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠⎝⎠
.
Therefore,
()
111 abbc ac ac
abc 3 52
abc bcab ca ca
⎛⎞⎛ ⎞ ⎛⎞
++ ++ =+ +++ ++≤+ +
⎜⎟⎜ ⎟ ⎜⎟
⎝⎠⎝ ⎠ ⎝⎠
.
On the other hand,
⎛⎞⎛ ⎞ ⎛⎞
≤≤⇒ − − ≤⇒ +≤ ⇒+≤
⎜⎟⎜ ⎟ ⎜⎟
⎝⎠⎝ ⎠ ⎝⎠
2
1a a1a a 5a ac5
12 0 1.
2c c2c c 2c ca2
.
Thus,
()

111
abc 10
abc
⎛⎞
+
+++≤
⎜⎟
⎝⎠
.
15.
()
a,b,c,d 0> ,
2222
22 22
4
abcdabcd
bcda
abcd
+
++
+++≥
.
Solution.
Applying the AM – GM Inequality we get
222 6
8
222 222
4
abc a 8a
3. 2. 2 8

bcd bcd
abcd
+++≥ =
,
222 6
8
222 222
4
bcd b 8b
3. 2. 2 8
cda cda
abcd
+++≥ =
,
222 6
8
222 222
4
cda c 8c
3. 2. 2 8
dab dab
abcd
+++≥ =
,
222 6
8
222 222
4
dab d 8d
3. 2. 2 8

abc abc
abcd
+++≥ =
.
Adding these four inequalities, we obtain
2222
22 22
444
a b c d abcd abcd abcd
68862
bcda
abcd abcd abcd
⎛⎞
+
++ +++ +++
⎛⎞⎛⎞⎛⎞
+++ +≥ = +
⎜⎟
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠
⎝⎠


4
444
abcd 4abcd abcd
62.68
abcd abcd abcd
+++ +++
⎛⎞ ⎛⎞

≥+=+
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
.
Thus,
2222
22 22
4
abcdabcd
bcda
abcd
+
++
+++≥
.
16.
()
0x2≤≤ ,
−+ +≤
33
4
4x x x x 3 3 .
Solution.
Applying the Cauchy – Buniakowski Inequality we get
(
)
()
()
⎡⎤
−+ + = − + + ≤ + − ++ =

⎣⎦
33 3 3 33
11
4x x x x 2 8x 2x 2 x x 2 4 8x 2x x x
22


()
()()
3
22 2
2
22
4
4
4
44
2x 9x 9x
16 6
.2x 9 x 33
23
222
⎛⎞
+− +−
⎜⎟
=−= =
⎜⎟
⎝⎠
.
17.

()
x ∈  ,
+
−≤2sinx 15 10 2cosx 6.
Solution.
Applying the Cauchy – Buniakowski Inequality we get
()
()()
⎡⎤
+− ≤+ +− = − +≤
⎢⎥
⎣⎦
2
2
2 sin x 15 10 2 cosx 1 5 2sin x 3 2 2 cosx 6 6 2 cosx 1 6
.
18.
()
222
x, y,z 0,x y z 1,n
+
>++=∈ ,
(
)
+
+
++≥
−−−
2n
2n 2n 2n

2n 1 2n 1
xyz
1x 1y 1z 2n
.
Solution.
We set
()
()
()
2n
ft t1 t , t 0,1=− ∈ . It is easy to show that
() ( ) () ()
()
()
+
⎛⎞
=− + = ⇔= ⇒ ≤ ⇒ − ≤
⎜⎟
+
+++
⎝⎠
2n 1 2n
2n 2n 2n
11 2n
f' t 1 2n 1 t , f' t 0 t f t f t 1 t
2n 1 2n 1 2n 1 2n 1

Since
0x,y,z1<<, thus

()
()
2n
2n
2n 1 2n 1
1
,
2n
x1 x
+
+
≥
−

()
(
)
2n
2n
2n 1 2n 1
1
,
2n
y1 y
+
+
≥
−

()

()
2n
2n
2n 1 2n 1
1
2n
z1 z
++
≥
−

Adding these three inequalities, we obtain
()()()
⎛⎞
⎛⎞
⎜⎟
+
+= + + ≥
⎜⎟
⎜⎟
−−−
−−−
⎝⎠
⎝⎠
222
2n 2n 2n
2n 2n 2n
xyz x y z
1x 1y 1z
x1 x y1 y z1 z

()
(
)
()
222
2n
2n
xyz2n12n1
2n 1 2n 1
2n 2n
++ + +
+
+
≥=
.
19.
()
a,b 0,a b 1>+<
,
+
+++≥
−− +
22
ab 1 9
ab
1a 1b ab 2
.
Solution.

We have,
22 2 2
ab 1 a b 1 111
ab 1a 1b 2 2
1a1b ab 1a 1b ab 1a1b ab
⎛⎞⎛ ⎞
⎛⎞
+ + ++ = +++ +++ − = + + −
⎜⎟⎜ ⎟
⎜⎟
−− + − − + −− +
⎝⎠
⎝⎠⎝ ⎠

Applying the Cauchy – Buniakowski Inequality we get
()
()()( )
2
111
111 9
1a1b ab 11a 11b 1.ab 2
++
⎛⎞
++ ≥ =
⎜⎟
−− + −+−++
⎝⎠
.
20.
()

a, b,c 0,abc 1>=,
22 22 22
1111
a2b3b2c3c2a32
+
+≤
++ ++ ++
.
Solution.
Applying the AM – GM Inequality we get
+≥
22
ab2ab
and
+
≥
2
b12b.
Therefore,
()
()
++≥ ++⇒ ≤
+
+++
22
22
11
a2b32abb1
a2b32abb1
.

Similarly,
()
≤
++ ++
22
11
b2c32bcc1
and
()
≤
+
+++
22
11
c2a32aca1
.
Adding these three inequalities, we obtain
⎛⎞
+
+≤ ++=
⎜⎟
++ ++ ++ ++ ++ ++
⎝⎠
22 22 22
1111111
a2b3b2c3c2a32abb1bcc1caa1


1 1 ab b 1 ab b 1 1
2 ab b 1 b 1 ab 1 ab b 2 ab b 1 2

++
⎛⎞
=++==
⎜⎟
++ ++ + + ++
⎝⎠
.
21.
()
22
x, y 0,x y 1>+=
,
() ()
⎛⎞
⎛⎞
+++++≥+
⎜⎟
⎜⎟
⎝⎠
⎝⎠
11
1x1 1y1 432
yx
.
Solution.
We have,
() ()
⎛⎞ ⎛ ⎞⎛⎞⎛⎞
⎛⎞⎛ ⎞
+ +++ +=++++++ ++

⎜⎟ ⎜ ⎟⎜⎟⎜⎟
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
⎝⎠ ⎝ ⎠⎝⎠⎝⎠
1111xy111
1x1 1y1 x y 2
y x 2x 2y y x 2 x y
.
Applying the AM – GM Inequality we get
11
x2, y2
2x 2y
+≥ +≥,
22
22
4
11 1 1 1 2
2
2x y x y
xy
xy
⎛⎞
+≥ = ≥ =
⎜⎟
+
⎝⎠
,
xy
2
yx

+≥.
Adding these four inequalities, we obtain
() ()
⎛⎞
⎛⎞
+++++≥+
⎜⎟
⎜⎟
⎝⎠
⎝⎠
11
1x1 1y1 432
yx
.
22.
()
0a,b,c1<≤
,
()()()
11
1a1b1c
abc 3
≥+− − −
++
.
Solution.
Applying the AM – GM Inequality we get
()()( )
()()( )()()( )
3

1b 1c 1bc
11b1c1bc1b1c1bc1
3
−+−+++
=≥−−++⇔−−++≤.
Therefore,
()()()
−−
−−−≤ ≤ ≤ −
++ ++ ++
1a 1a 1 1
1a1b1c
1bc abc abc 3

()()()
⇒≥+−−−
+
+
11
1a1b1c
abc 3
.
23.
()
22 2
xyz0,323x z 164y≥≥> − = = − ,
+
++≤
32 3 16
xy yz zx

5
.
Solution.
We have
22
22
16 z 32 z
y, x
43
−−
==,
−
≥⇒ ≥ ⇒≤
2
2
16 z 4
yz z z
4
5
.
On the other hand,
−− −
−= − = ≤⇒≤ ⇒≤
222
22 2
32 z 48 3z 5z 16
x3y 0xy3 xy 3y
3412
.
We have

()
2
222
x3x 3
xz 3 .z z y z
23 2
3
⎛⎞
⎛⎞
≤≤+≤+
⎜⎟
⎜⎟
⎝⎠
⎝⎠
and
()
22
1
yz y z
2
≤+.
Adding these two inequalities, we obtain
()
()()
2
22222 2
31 3116z31
x
yy
zzx 3

yy
z
y
z3 z
22 20242
⎛⎞⎛⎞
⎛⎞
−+
++ ≤ + + + + = + + +
⎜⎟⎜⎟
⎜⎟
⎜⎟⎜⎟
⎝⎠
⎝⎠⎝⎠


() ()
2
33 331632316
23 3 1 z 23 3 1 .
8855
⎛⎞ ⎛⎞
+
++
=++ ≤++ =
⎜⎟ ⎜⎟
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
.
24.

+
π
⎛⎞
<< ∈
⎜⎟
⎝⎠
0x ,n
2
,
n2 n2
nn
sin x cos x
1
cos x sin x
++
+
≥ .
Solution.
We have,

(
)
(
)
−
−≥
2n 2n 2 2
sin x cos x sin x cos x 0

()

(
)
⇔− − ≥
nn22
tg x cot g x sin x cos x 0

n2 n2 n2 n2
tg xsin x cot g x cos x cot g x sin x tg x cos x⇔+ ≥ +.
Thus,
()( )
n2 n2
nn22 nn
nn
sin x cos x 1 1
tgx cotgx sin x cos x .2. tgx.cotgx 1
cos x sin x 2 2
++
+≥+ +≥ =.
25.
1
0 a,b,c,0 x,y,z ,a b c x y z 1
2
⎛⎞
<≤≤++=++=
⎜⎟
⎝⎠
,
ax by cz 8abc
+
+≥

.
Solution.
Without loss of generality, we can assume that
abc
≤
≤
.
We have,
()() ()
−
≤+ =− ⇒ ≤ − ≤
22 2
1c
4ab a b 1 c 8abc 2c 1 c
2
.
() ()
11111ab 1c
0 x, y, z cz c x c y a x b y ax by ax by
222222 2
+−
⎛⎞⎛⎞⎛⎞⎛⎞
≤ ≤⇒= −+ −≥ −+ −= − + = − +
⎜⎟⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠⎝⎠
.
Thus,
−
++≥ ≥
1c

ax by cz 8abc
2
.
26.
()
33
x, y 0,x y 1≥+=,
()
+≤+
5
5
6
x2y 122.
Solution.
We set
5
5
1
, 2 2
122
α= β= α
+
.
Applying the AM – GM Inequality we get
3
6335
6
5
x5
x5 6x x

6
+
α
+α≥ α⇒ ≤
α
,
3
335
6
5
6
y5
x5 6y 2y
3
+
β
+β≥ β⇒ ≤
β
.
Adding these two inequalities, we obtain
()
(
)
()
33
5
5
6
6
5

xy5
x2y 122
6
++α+β
+≤ =+
α
.
27.
3x 4y 2z
x, y,z 0, 2
x1 y1 z1
⎛⎞
>++=
⎜⎟
+++
⎝⎠
,
342
9
1
xyz
8
≤
.
Solution.
Applying the AM – GM Inequality we get

=− = + + =
+++++
1 x 2x 4y 2z

1
x1 x1 x1 y1 z1


()()()
=+++++++≥
++++++++
+++
242
8
242
xxyyyyzz xyz
8
x1 x1 y1 y1 y1 y1 z1 z1
x1 y1 z1
.
Similarly,
()()()
≥
+
+++
332
8
332
1yxz
8
1y
y1 x1 z1
and
()()()

≥
+
+++
34
8
342
1xyz
8
1z
x1 y1 z1
.
Thus,
()()() ()()()
4
32
24 32 16 3 4 2
99
8
24 32 16 3 4 2
111 xyz xyz
88
1x 1y 1z
x1 y1 z1 x1 y1 z1
⎛⎞
⎛⎞ ⎛⎞
≥=
⎜⎟
⎜⎟ ⎜⎟
+++
⎝⎠ ⎝⎠

+++ +++
⎝⎠

or
342
9
1
xyz
8
≤
.
28.
()
a,b,c 0> ,
()
333
333
111 3bccaab
abc
abc 2a b c
+
++
⎛⎞⎛ ⎞
++ + + ≥ + +
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
.
Solution.
We have,
()

()
(
)
(
)
333
2a b c aba b bcb c cac a++ ≥ ++ ++ + and
333
111 3
abcabc
++≥ .
Therefore,
()
333
333
111 3bccaab
abc
abc 2a b c
+
++
⎛⎞⎛ ⎞
++ ++ ≥ + +
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠

29.
()
x ∈  ,
()
++ ≥

4
4
16cos x 3 768 2048cosx .
Solution.
Applying the AM – GM Inequality we get
()( )
(
)
+= ++++ ≥ + =
4
44
44 4
4
16cos x 3 16cos x 1 1 1 768 4 16cos x 768

443
4
4096 cos x 256 256 256 4 4096cos x.256 2048 cosx 2048cosx=+++≥ =≥.
30.
()
x ∈ 
,
()
()
++
≤
≤
+
8
4

4
2
1x 16x
1
17
8
1x
.
Solution.
If a, b are two real numbers, then
(
)
()
2
4
22
44
ab
ab
ab
28
+
+
+≥ ≥ .
Thus,
()
()
() ()
()
()

()
⎡⎤
++−
+
++
⎣⎦
=
≥=
+++
4
24
4
8
2
4
444
222
1x 2x
1x
1x 16x
1
8
1x 1x 81x
.
On the other hand,
()
() ()
4
28
4

2
1x 1x
1x
216
⎡⎤
++
+≥ ≥
⎢⎥
⎢⎥
⎣⎦
and
()()
()
4
44
2
22 4
1x 1x 2x 2x 1 x 2x 16x
⎡⎤
+=+−+ =−+ ≥
⎣⎦
.
Therefore,
()
()
++
≤
+=
+
8

4
4
2
1x 16x
16 1 17
1x

31.
()
a,b,c 0>
,
22 2222 222
ab bc ca abc
3
ab bc ca abc
+++ ++
++≤
+++ ++
.
Solution.

22 2222 222
ab bc ca abc
3
ab bc ca abc
+
++ ++
++≤
+++ ++

()
()
⎡⎤
+++
⇔++ + + ≤ ++
⎢⎥
+++
⎣⎦
22 2222
222
ab bc ca
abc 3a b c
ab bc ca


()()
(
)
22 22 22
222
ca b bc a ab c
abc
ab ca bc
+++
⇔++≤++
+++


()

(
)
(
)
22 22 22
222
ca b bc a ab c
cba0
ab ca bc
+++
⇔− +− +− ≥
+++


() ()
(
)
(
)
(
) ()
ac c a bc c b ab b a bc b c ab a b ac a c
0
ab ab ca ca bc bc
−− −−−−
⇔+++++≥
++++ ++


()

()()
()
()()
()
()()
222
ac c a bc c b ab b a
0
abbc abac acbc
−−−
⇔++≥
++ ++ ++
.
32.
()
a,b,c 0,n ,n 2>∈ ≥ ,
n
nnn
abcn
n1
bc ca ab n1
++ > −
+++−
.
Solution.
Applying the AM – GM Inequality we get
()()
()
()( )
−

+
⎛⎞
−++++
⎜⎟
+− −++
⎝⎠
≤=⇒≥−
+
−++

n
n1
n
n
ab
n1 11 1
abn1 n1abc
c
cn c
n1
c n nc ab n1 abc
.
Similarly,
≥−
+− ++
n
n
bn b
n1
ca n1 abc

and
≥−
+
−++
n
n
an a
n1
bc n1 abc
.
Adding these three inequalities, we obtain
n
nnn
abcn
n1
bc ca ab n1
+
+≥ −
+++−
.
Equality holds if and only if
(
)( )
()()
()()
()( )
()
−+=
⎧
⎪

−+=⇒−++=++⇒=
⎨
⎪
−+=
⎩
n1ab c
3
n1ca b 2n1abc abc n !
2
n1bc a
.
Therefore,
n
nnn
abcn
n1
bc ca ab n1
+
+> −
+++−
.
33.
()
x, y, z 0, x y z 1, n ,n 2≥++=∈≥ ,
()
+
++≤
+
n
nnn

n1
n
xy yz zx
n1
.
Solution.
Without loss of generality, we can assume that
{
}
xmaxx,
y
,z=
.
Since
zx≤ , thus
nnn2n1
zx zx, zx zx
−
≤≤
.
On the other hand,
−
−
>⇒ ≥ ⇒ ≥
n1 1 n1 z
n1 z
n2 n 2
.
We have,
−

−−
++≤+ + + ≤+ ++ =
n2n1
nnn nn1 n n nn1
11 zxzx
xy yz zx xy x yz zx zx xy x yz
22 22


() ()
n1 n1 n
zn1xxxxzn1
x xzy x xzy z n .y z
2nnnnnn
−−
−
⎡+−⎤
⎛⎞ ⎛⎞ ⎛⎞
=++≤++ ≤ +
⎜⎟ ⎜⎟ ⎜⎟
⎢
⎥
⎝⎠ ⎝⎠ ⎝⎠
⎣
⎦


()
()
()

()
n1
n1
n
nn
n1 n1
xxz n1
n1 y z
xyz
n
nn n
nn.
n1
n1 n1
+
+
+
+
+−
⎡⎤
−+ ++
⎢⎥
++
≤==
⎢⎥
+
++
⎢⎥
⎣⎦
.

34.
()
x, y,z 0> ,
( )()()()
444
3
16x
y
zx
y
z3x
yy
zzx++ ≤ + + +
.
Solution.
Applying the AM – GM Inequality we get
()()()()()()
+++=+++++++++
111
xyyzzx xyzxy xyzxy xyzxy
333


()()()
+
++ + ++ + ++ + + ≥
22
111
yz x y z yz x y z yz x y z xz zx
333

()( ) ()( )
++ ++
≥=
66 33
8
4
6
xyz xyz xyz xyz
88
327
.
Thus,
( )()()()
444
3
16x
y
zx
y
z3x
yy
zzx++ ≤ + + +
.
35.
x, y,z ,
62
⎛ππ⎞
⎡⎤

∈
⎜⎟
⎢⎥
⎣⎦
⎝⎠
,
2
sin x siny siny sinz sinz sinx 1
1
sinz sin x sin y
2
−−−
⎛⎞
++ ≤−
⎜⎟
⎝⎠
.
Solution.
We set
a sin x, b sin y, c sin z=== and observe that
1
a, b, c ,1
2
⎡
⎤
∈
⎢
⎥
⎣
⎦

.
The inequality is equivalent to
(
)
(
)
(
)
22
abbcca
ab bc ca 1 1
11
cab abc
22
−−−
−−−
⎛⎞ ⎛⎞
++ ≤− ⇔ ≤−
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
.
Without loss of generality, we can assume that
1
abc1
2
≤
≤≤≤.
We set
ab
u, v

cc
== and observe that
1
uv1
2
≤
≤≤.
We will prove that
()
(
)
(
)
2
vu1u1v
1
1
uv
2
−−−
⎛⎞
≤−
⎜⎟
⎝⎠
.
We have
()()()
()
2
11

v11v
vu1u1v
111 1 1
22
1v 12v. 1
1
uv 2 2v 2 2v
2
v
2
⎛⎞⎛⎞
−−−
⎜⎟⎜⎟
−−−
⎛⎞
⎝⎠⎝⎠
≤ =+−− ≤+− = −
⎜⎟
⎝⎠
.
36.
()
(
)
()
x, y, z, t , x y z t xy 88 0∈++++=
,
+++ ≥
222 2
x 9y 6z 24t 352 .

Solution.
Since
()()
xyzt xy880++++=, thus
(
)
(
)
4 x y z t 4xy 352 0
+
++ + =.
We have
(
)
(
)
+++ ++ ++ =+++ +++++ =
222 2 222 2
x9
y
6z 24t 4 x
y
zt 4x
y
x9
y
6z 24t 4xz 4xt 4
y
z4
y

t4x
y


()()
=+ +++ ++ + − ++− + +− + =
2
2222222
x 4x z y t 4 z y t 4y 4yz z z 8zt 16t y 4yt 4t

()()()()
2222
x2z
y
t2
y
zz4t
y
2t 0
⎡⎤
=+ ++ + − +− +− ≥
⎣⎦
.
Therefore,
+++ ≥
222 2
x 9y 6z 24t 352 .

37.
()

x, y,z 0,xyz 1>=,
+
+≥
++ ++ ++
222
33 3
xyz
3
xyyz yzzx zxxy
.
Solution.
We set
=++
++ ++ ++
222
33 3
xyz
A
x
yy
z
y
zzx zxx
y
.
We have
=++=++
++ ++ ++ ++ ++ ++
333333
232323222222

xyzxyz
A
xyzyzxyzyzxyzxzxyzxyxyyzyz zxzx
.
Applying the Cauchy – Buniakowski Inequality we get
()
(
)
(
)
(
)
2
222222222
A. x x xy y y y yz z z z xz x x y z
⎡⎤
++ + ++ + ++ ≥ ++
⎣⎦


()
()
(
)
2
222 222
Ax y z x y z x y z⇔++ ++≥++.
Thus,
++ ++
≥≥≥=

++
222
3
3xyz
xyz xyz
A3
xyz 3 3
.
38.
()
x, y,z 0,xyz 1>=,
+
+≤
++ ++ ++
22 22 22
22 7 7 22 7 7 22 7 7
xy yz zx
1
xy x y yz y z zx z x
.
Solution.
If x, y are positive real numbers, then
(
)
77 33
x
y
x
y
x

y
+
≥+
.
Thus,
() () () ()
22 22
22 7 7 22 33
xy xy 1 1 z z
x
y
x
y
x
y
x
y
x
y
1x
y
x
y
x
y
zx
y
x
y
x

y
zx
y
zx
y
z
≤====
++ + + ++ ++ ++ ++
.
Similarly,
≤
++ ++
22
22 7 7
yz x
yz y z x y z
and
≤
+
+++
22
22 7 7
zx y
zx z x x y z
.
Adding these three inequalities, we obtain
++
++≤=
++ ++ ++ ++
22 22 22

22 7 7 22 7 7 22 7 7
xy yz zx x y z
1
xy x y yz y z zx z x x y z
.
39.
()
a,b,c 0,a b c 1>++=,
++≤+
+++
ababc33
1
abc bca cab 4
.
Solution.
We set
ababc
T
abc bca cab
=++
+
++
.
We have
ab
ababc11
c
T
bc ca ab
abc bcacab

111
abc
=++=++
+++
+++
.
We set
==<<π
22
bc A ca B
tan , tan , 0 A, B
a2b2
. Since
ab ca bc ab ca bc
1abc . . .
cb ac ba
=++= + +

thus,
−
+
⎛⎞
==
⎜⎟
⎝⎠
+
AB
1tan tan
ab A B
22

cot g
AB
c2
tan tan
22
.
Therefore,
AB
22
+π
<
and
()()
()
ab C
tg , C A B 0,
c2
=
=π− + ∈ π .
We obtain
()
=++=++=+++
+++
22
222
C
tan
11 ABsinC1
2
T cos cos 1 cosA cosB sinC

ABC
2222
1tan 1tan 1tan
222
.
On the other hand,
CC
AB AB
33
cosA cosB sin C sin 2cos .cos 2sin .cos
322 22
ππ
+−
π+−
+++= +


CABC
AB
33
2 cos 2.cos 4cos 4 cos 2 3
22 4 6
π
π
−++−
+π
≤+ ≤ ==
.
Thus,
1333

T1 2.3 1
224
⎛⎞
≤+ − =+
⎜⎟
⎜⎟
⎝⎠
.
40.
()
x, y, z 0, x y z 2007>++= ,
++≥
20 20 20
9
11 11 11
xyz
3.669
yzx
.
Solution.
Applying the AM – GM Inequality we get
+ + = +++ ++ + + + ≥ =


20 20 20
11 8
20
11 8 11 8 11 8
8
11

xx x
11y 8.669 y y y 669 669 669 20 .y .669 20x
y 669 y 669 y .669
.
Similarly,
++ ≥
20
11 8
y
11z 8.669 20y
z.669
and
++ ≥
20
11 8
z
11x 8.669 20z
x .669
.
Adding these three inequalities, we obtain
()
20 20 20
89
11 11 11
xyz
9 x y z 3.8.669 .669 3.669
yzx
++≥ ++− =⎡⎤
⎣⎦

41.
()
a,b,c 0> ,
33 3 3 33
222
5b a 5c b 5a c
abc
ab 3b bc 3c ca 3a
−−−
+
+≤++
+++
.
Solution.
Since
()
33
ababab+≥ +
, thus
(
)
(
)
(
)
(
)
33 3 2 2
a5b abab6b baab6b ba3ba2b−≥ +−= +− =+ −

()
(
)
⇒−≤+ −
33 2
5b a ab3b 2ba


−
⇒≤−
+
33
2
5b a
2b a
ab 3b
.
Similarly,
−
≤
−
+
33
2
5c b
2c b
bc 3c
and
−

≤
−
+
33
2
5a c
2a c
ca 3a
.
Adding these three inequalities, we obtain
33 3 3 33
222
5b a 5c b 5a c
abc
ab 3b bc 3c ca 3a
−−−
+
+≤++
+++
.
42.
()
(
)
()
>∈+++=
22 22
a,b 0,x,y,z,t ,a x y b z t 1 ,
()()
+

++≤
ab
xzyt
2ab
.
Solution.
We have
()()
22 22
22 22
xzyt 1
ax y bz t 1
babaab
++ +=⇔+++=
.
Applying the Cauchy – Buniakowski Inequality we get
() ()
⎛⎞
++≥+
⎜⎟
⎝⎠
22
2
xz
ba xz
ba
and
() ()
⎛⎞
+

+≥+
⎜⎟
⎝⎠
22
2
yt
ba yt
ba
.
Adding these two inequalities, we obtain
()()()
2222
22
xzyt ba
xz yt ba
baba ab
⎛⎞
+
+++≤+ +++ =
⎜⎟
⎝⎠

()()
()()
+++
+
⇒+ +≤ ≤
22
xz yt
ab

xzyt
22ab

43.
()
333 222
x, y,z 1,x y z x y z>− + + ≥ + + ,
555 222
xyzxyz
+
+≥++.
Solution.
Since x1>− , thus
()( )
− + ≥⇔ − +≥⇔ ≥ −
2
3532
x1 x2 0 x 3x2 0 x 3x 2x.
Similarly,
≥−
532
y3y2y and ≥−
532
z3z2z.
Adding these three inequalities, we obtain
(
)
(
)
(

)
(
)
(
)
5 55 3 33 2 22 333 3 33 2 22 333
xyz3xyz 2xyz xyz 2xyz xyz xyz
⎡⎤
++≥ ++− ++=+++ ++−++ ≥++
⎣⎦
.
44.
()
,, ,sin sin sin 2αβγ∈ α+ β+ γ≥
,
cos cos cos 5α+ β+ γ ≤ .
Solution.
Applying the Cauchy – Buniakowski Inequality we get
(
)
222 222
cos cos cos 3. cos cos cos 3. 3 sin sin sinα+ β+ γ ≤ α+ β+ γ = − α+ β+ γ


()
2
sin sin sin
4
3. 3 3. 3 5
33

α+ β+ γ
≤− ≤−=
.
45.
()
>++=a, b, c 0, a b c 6 ,
+++++≥
+++
222
111317
abc
bc ca ab 2
.
Solution.
First solution.
Applying the Cauchy – Buniakowski Inequality we get
⎛⎞
⎛⎞
⎜⎟
+=+=++++≥ ++++ = +
⎜⎟
⎜⎟
++ +
++
⎝⎠
⎜⎟
⎝⎠


2

2
2222
2
16
16
116a 1aa a 1 1aa a 1 1 1
a 4a
b c 16 b c 16 16 16 b c 17 4 4 4 17
bc bc
.
2
11 1
a4a
bc
17 b c
⎛⎞
⇒+ ≥ +
⎜⎟
+
+
⎝⎠
.
Similarly,
⎛⎞
⇒+ ≥ +
⎜⎟
+
+
⎝⎠
2

11 1
b4b
ca
17 c a
and
⎛⎞
+≥ +
⎜⎟
+
+
⎝⎠
2
11 1
c4c
ab
17 a b
.
Adding these three inequalities, we obtain
()
⎡
⎤
⎛⎞
+++++≥ +++ + + ≥
⎢
⎥
⎜⎟
+++
+++
⎝⎠
⎣

⎦
222
1111 111
abc 4abc
bc ca ab
17 a b b c c a


()()()
()()()
⎛⎞
⎛⎞
⎜⎟
⎜⎟
⎜⎟
≥+ ≥+ =
⎜⎟
⎜⎟
+++++
⎜⎟
+++
⎜⎟
⎝⎠
⎝⎠
3
131 3317
24 24
2
17 17
ab bc ca

abbcca
3
.
Second solution.
Applying the Minkowski Inequality and the AM – GM Inequality we get
()
⎛⎞
+++++≥+++ + + ≥
⎜⎟
+++
+++
⎝⎠
2
2
222
111 111
abc abc
bc ca ab
ab bc ca

()()()
()()()
2
2
3
33317
36 36
2
ab bc ca
abbcca

3
⎛⎞
⎛⎞
⎜⎟
⎜⎟
⎜⎟
≥+ ≥+ =
⎜⎟
⎜⎟
+++++
⎜⎟
+++
⎜⎟
⎝⎠
⎝⎠
.
46.
()
2 2 22 22 222
x, y, z , x 2y 2x z y z 3x y z 9∈++ ++ =
,
≥−xyz 1.
Solution.
Applying the AM – GM Inequality we get
22222222222222222222 121212
9
9xyyxzxzxzyzxyzxyzxyz9xyz=+++++++ + + ≥ .
Thus,
≥⇒≥−1xyz xyz 1
47.

()
x, y,z,t 0,xyzt 1>=,
()()()()
+++ ≥
++ ++ ++ + +
3333
111 14
xyzztty yxzzttx zxttyyx txyyzzx 3
.
Solution.
We set
1111
a , b , c , d
xyzt
====
and observe that abcd 1
=
. The inequality can rewrite
+
++≥
++ ++ ++ ++
2222
abcd4
bcd cda dababc 3
.
Applying the Cauchy – Buniakowski Inequality we get
()
()
+++
+++

+++≥ = ≥ =
++ ++ ++ ++ +++
2
2222
4
abcd
abcd abcd4abcd4
bcd cdadababc 3abcd 3 3 3
.
48.
()
+
∈>+++≥
12 k 1 2 k
k,n ,a , a , , a 0,a a a k ,
++ +
+++
≤
+++
nn n
12 k
n1 n1 n1
12 k
a a a
1
aa a
.
Solution.
If a is a positive real numbers then
()

(
)
nn1n
a1a10 a a a1
+
−−≥⇔−≥−.
Therefore,
()
(
)
(
)
n1 n1 n1 n n n
12 k 12 k 12 k
a a a a a a a a a k 0
++ +
+++ −+++≥+++−≥.
49.
()
a,b,c 0,abc a c b>++=,
−+≤
+++
222
22310
a1b1c13
.
Solution.
We have
ac
abc a c b ac 1

bb
+
+= ⇔ + + =
.
Since
a, b, c 0> , thus there exist
(
)
A, B, C 0,
∈
π
such that
ABC++=π
and
===
A1 B C
a tan , tan , c tan
2b 2 2
.
Therefore,
−
−
−+=− + +≤ +≤+=
+++
22
222
223 C CAB1AB110
3sin 2sin cos 3 cos 3 3
a1b1c1 2 2 2 3 3 3 3
.

50.
()
∈+−−≤
22
x,y ,x y 2x 2y 0 ,
+
≤+2x y 10 3 .
Solution.
Applying the Cauchy – Buniakowski Inequality we get
()() ()()
+−= −+−≤ − +− ≤ =
22
2x y 3 2 x 1 y 1 5. x 1 y 1 5. 2 10 or +≤ +2x y 10 3 .
51.
3
0x,y,z1,xyz
2
⎛⎞
≤≤++=
⎜⎟
⎝⎠
,
222
5
xyz
4
+
+≤.
Solution.
It is easy to show that

()
1
x, y,z 1, ,0
2
⎛⎞
⎜⎟
⎝⎠
≺ and
(
)
2
ft t
=
is convex on
[
]
0,
+
∞
.
Applying the Karamata Inequality we get
() () () () ()
1
fx fy fz f0 f f1
2
⎛⎞
++≤+ +
⎜⎟
⎝⎠
or

222
5
xyz
4
+
+≤.
52.
()
a,b,c 0>
,
ab bc ca c a b
2
cab abbcac
⎛⎞
+++
++≥ ++
⎜⎟
⎜⎟
+
++
⎝⎠
.
Solution.
If x, y are two positive real numbers, then
()
+≤ + +≥
+
11 4
xy2xy,
xyxy

.
Thus,
ab bc ca 1 a b 1 b c 1 c c
cab
2c c 2a a 2b b
⎛⎞⎛⎞⎛⎞
+++
++≥+++++
⎜⎟⎜⎟⎜⎟
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠


a1 1 b1 1 c1 1 22a 22b 22c
2b c 2a c 2a b bc ac ab
⎛⎞⎛⎞⎛⎞
=+++++≥++
⎜⎟⎜⎟⎜⎟
+++
⎝⎠⎝⎠⎝⎠


() () ()
22a 22b 22c a b c
2
bc ac ab
2b c 2a c 2a b
⎛⎞
≥++=++
⎜⎟

⎜⎟
+++
+++
⎝⎠
.
53.
()
a,b,c 0,ab bc ca 1>++=
,
10 3
abcabc
9
+++ ≥ .
Solution.
We have
()( )
2
abc 3abbcca 3 abc 3++ ≥ + + =⇒++≥ .
Applying the AM – GM Inequality we get
()()()
3
3
abc 3
1a1b1c 1 1
33
⎛⎞
⎡++⎤
⎛⎞
−−−≤− ≤−
⎜⎟

⎜⎟
⎢⎥
⎜⎟
⎝⎠
⎣⎦
⎝⎠


⎛⎞
⇔+ + + −−−− ≤ −
⎜⎟
⎜⎟
⎝⎠
3
3
1abbccaabcabc 1
3


10 3
abcabc
9
⇔+++ ≥ .
54.
()
12 n
0 a , a , , a 1, n 2≤≤≥,