50 amc lectures problems book 1 (25) am-gm ...50 amc lectures problems book 1 (25) am-gm...

50 AMC Lectures Problems Book 1 (25) AM-GM Inequalities

238

PROBLEMS

Problem 1: (1958 AMC) For values of x less than 1 but greater than – 4, the expression

22

222

x

xx has:

(A) no maximum or minimum value

(B) a minimum value of +1

(C) a maximum value of +1

(D) a minimum value of –1

(E) a maximum value of –1

Problem 2: If xyz = 27, x, y, and z are positive, find the minimum value of x + y + z.

Problem 3: x and y are real positive numbers with x + 2y = 1. Find the smallest value for

yx

11 .

Problem 4: Let 5a , find the smallest value of 5

4

aa .

Problem 5: Show that (x + y)(y + z) 2 if xyz (x + y + z) = 1, where x, y, z are positive

numbers.

Problem 6: Find the smallest value of .)(

1

bbaa

a > b > 0.

Problem 7: Find the smallest value of x + y if 1y

b

x

a. x, y, a, and b are positive real

numbers.

Problem 8: Find the greatest possible value of 13 a + 13 b + 13 c if a + b + c

= 1.


239

Problem 9: Find the smallest value for

bb

aa

11 if a + b = 1, a > 0, b > 0.

(A) 6 (B) 4

25 (C)

4

27 (D) None of them.

Problem 10: (1979 AMC) For each positive number x, let

.11

211

)(

3

3

3

6

6

6

xx

xx

xx

xx

xf The minimum value of f(x) is

(A) 1 (B) 2 (C) 3 (D) 4 (E) 6

Problem 11: Show that cb

a

+

ac

b

+

ba

c

> 2. a, b, c > 0,

Problem 12: Show that 8

144 ba if a + b = 1.

Problem 13: The smallest value of cdbdbcadacabdcba 2222 is 10

if abcd = 1. Prove it for positive numbers a, b, c, and d.

Problem 14: Farmer Bob has 96 square inches of wrapping paper. Find the volume of the

largest rectangular box he can wrap with the paper.

Problem 15: (1977 AMC) Find the smallest integer n such that

)()( 4442222 zyxnzyx for all real numbers x, y, and z.

(A) 2 (B) 3 (C) 4 (D) 6 (E) There is no such integer n.

Problem 16: (1975 AMC) Which of the following inequalities are satisfied for all real

numbers a, b, c, x, y, z which satisfy the conditions x < a, y < b, and z < c?

I. xy + yz + zx < ab + bc + ca

II. x2 + y

2 + z

2 < a

2 + b

2 + c

2

III. xyz < abc

(A) None are satisfied. (B) I only (C) II only (D) III only (E) All are satisfied.


240


6

cb

a +

22

6

ca

b + cabcab

ba

c

22

6

.

Problem 18: Show that A

1 +

B

1 +

91

Cif A, B, C are three interior angles of ABC.

Problem 19: Show that acb

a

+

bac

b

+

cba

c

≥ 3

if a, b, c are three sides of ABC.

Problem 20: Prove that 132222 abccba if a + b + c = 1. a, b, c are positive

numbers.

Problem 21: If a, b, c R ，show that: )(2222222 cbaaccbba .


11

1

yxif x +y = 1 and x and y are positive numbers.

Problem 23: If a, b, and c are positive numbers, show that ：

)3

(3)2

(2 3 abccba

abba

Problem 24: ),0( cba 、、 . 1 cba . Show that 3 cba .

Problem 25: For a > b > c, show that .411

cacbba

Problem 26: If a, b, and c are positive integers less than 1, show that not all of (1－a)b,

(1－b)c, and (1－c)a greater than1/4.


241

SOLUTIONS:

Problem 1: Solution: (E).

1

22

2

1 2

x

xxy = ]

1

11[

2

1

xx = ]

1

1)1[(

2

1

xx

Note that – 4 < x < 1, so 1 – x > 0.

By AM–GM: 2)1

1)(1(2

1

1)1(

xx

xx .

Equality occurs when x

x

1

1)1( . Solving we get x = 0. Since y is negative 1,

y is a maximum when x = 0.

Problem 2: Solution: 9.

From the AM–GM inequality, we have:

327)(3

33

xyzzyx

. So 9 zyx .

The minimum value of x + y + z is 9. This is true when x = y = z = 3.

Problem 3: Solution: 223 .

Method 1:

yx

11 =

y

x

x

y

y

yx

x

yx

23

22.

Applying AM-GM yields

222

22

y

x

x

y

y

x

x

y. Substituting this into the above equation yields

yx

11 223 .

This value can be achieved when y

x

x

y

2, which gives 22x and

2

12 y .


242

Method 2:

We are given that x + 2y = 1. Applying Cauchy’s inequality yields:

yx

11 = 223

2

)21(

2

21 2

yxyx.

So the smallest value for yx

11 is 223 . This value can be achieved by letting

22x and 2

12 y .


Applying AM-GM yields

5

4

aa = 9545

5

4)5(25

5

45

aa

aa .

The smallest value is 9, which can be achieved when 5

45

aa or a = 7.

Problem 5: Solution:

(x + y)(y + z) = xz + y(x + y + z) ≥ 2 )( zyxxyz = 2.


)()(

1ba

bbaa

+ ,3

)(

1

bbab

The smallest value 3 is achieved when a = 2, b = 1.


From a > 0, b > 0, x > 0, y > 0 and 1y

b

x

a, we get

x > a, y > b (otherwise y

b

x

a would be greater than 1) and 1

xy

bxay.

Therefore we can write x – a > 0, y – b > 0, and xy – ay – bx = 0 (1)

Add ab to both sides of (1): xy – ay – bx + ab = ab (2)

Factoring (1): (x – a)(y – b) = ab.

By AM – GM:


243

x + y = (x – a) + (y – b) + (a + b) ≥ )())((2 babyax

= baab 2 = 2)( ba .

Thus ( 2)() bayx (3)

We know that x and y satisfy 1y

b

x

a and equality occurs in (3) when x – a = y – b .

Therefore we have

.

))((

byax

abbyax

Solving we get abax and abby .

Then we have abayx min)( + abb = 2)( ba .


Let ,13 ax ,13 by ,13 cz t = x + y + z.

Then zxyzxyzyxt 2222222 .

By AM – GM (25.14), we have t2 ≤ 3(x

2 + y

2 + z

2) = 18.

Therefore 23t . Equality occurs when a = b = c = 3

1.


bb

aa

11=

b

a

a

b

abab

12

16

15

16

12

1

ababab

abab

ab16

15

2

5

2)2

(16

15

2

5

ba =

4

25.

Note: The following method is not working:

bb

aa

11=

b

a

a

b

ab

ab

1

1

422 .

This is because the condition ab

ab 1

1 or ab = 1 will not be true. Considering ab = 1 and

a + b = 1 we get (a – b)2 = – 1.

Problem 10: Solution: (E).

By observing that ,1

21

6

6

2

3

3

xx

xx

one sees that


244

f (x) =

31

xx

3

3 1

xx = 3

xx

1 ≥ 3 × 2 = 6.

f(x) = 3

xx

1has a minimum value of 6, which is taken on at x = 1.


Since ,2

cbacba

so cb

a

= ,

2

2

cba

a

cba

a

cba

a

similarly ac

b

,

2

cba

b

ba

c

,

2

cba

c

Add them together we are done.


By AM – GM, 4

1

2

1

22

2222

baba.

16

1

4

1

222

22

222244

bababa.

Therefore 8

1

16

1244 ba .

Problem 13: Proof:

cdbdbcadacabdcba 2222

= (a – b)2 + (c – d)

2 + 3ab + ac + ad + bc + bd + 3cd

≥ 3ab + ac + ad + bc + bd + 3cd = 3(ab + cd) + (ac + bd) + (ad + bc)

= 3

abab

1+

acac

1+

adad

1 ≥ 3 × 2 + 2 + 2 = 10.


Let the box have dimensions x, y, and z. The amount of wrapping paper should equal the

surface area of the box, or


245

48)( 96)(2 zxyzxyzxyzxy

From the AM-GM inequality, we have

3 23 )())()((3

xyzzxyzxyzxyzxy

, that is 3 2)(16 xyz .

So the maximum value of xyz, the volume of the box, is 64. This value can be achieved

by letting x = y = z = 4.

Problem 15: Solution: (B).

Let a = x2, b = y

2 and c = z

2. By AMGM, a

2 + b

2 ≥ 2ab, we see that

(a + b + c)2 = a

2 + b

2 + c

2 + 2ab + 2ac + 2bc

≤ a2 + b

2 + c

2 + (a

2 + b

2) + (a

2 + c

2) + (b

2 + c

2)

= 3(a2 + b

2 + c

2).

Therefore n ≤ 3. Choosing a = b = c > 0 shows n is not less than three.

Problem 16: Solution: (A).

None of the inequalities are satisfied if a, b, c, x, y, z are chosen to be 1, 1, – 1, 0, 0, –10,

respectively.


We know that x2 + y

2 + z

2 ≥ xy + xz + yz.

Therefore we can have 2

3

bc

a+

23

ac

b+

2

332

3

abc

ba

ab

a

+

bca

cb2

33

+ cab

ca2

33

=

2

c

ab+

2

a

bc+

ca

cab

b

ac 22

+

ab

abc2

+ bc

bca2

= a2 + b

2 + c

2 ≥ ab + bc + ac

Therefore 22

6

cb

a +

22

6

ca

b + cabcab

ba

c

22

6

.


Method 1:

We know that A + B + C = π and A > 0, B > 0, C > 0.

Therefore

A

1 +

B

1 + )(

11

C

CBA

B

CBA

A

CBA

CBAC

= )]()()(3[1

)(1

C

A

A

C

C

B

B

C

B

A

A

B

C

C

C

B

C

A

B

C

B

B

B

A

A

C

A

B

A

A


246

9)2223(

1 .

Method 2:

We know that A + B + C = π and A > 0, B > 0, C > 0.

By AM – GM, A

1 +

B

1 +

33

313

1

ABCABCC

We have 33 ABCCBA

313

ABC

.

Therefore

9111

CBA.


Method 1:

We have x > 0, y > 0, z > 0. By AMGM (25.17), 9111

)(

zyxzyx .

We can write a + b + c = b + c – a + c + a – b + a + b – c

Therefore 9111

)(

cbabacacbcba .

Method 2:

Let 2x = b + c – a, 2y = a + c – b, 2z = a + b – c with x, y, z > 0.

Then a = y + z, b = z + x, c = x + y.

Therefore acb

a

+

bac

b

+

cba

c

=

z

yx

y

xz

x

zy

222

= 3)222(2

1)(

2

1

z

y

y

z

z

x

x

z

y

x

x

y.


Note that ,2 22222 cabcbba

,2 22222 abcaccb

,2 22222 bcabaac

So we have

,222222222 bcaabccabaccbba

so (ab + bc + ca)2 ≥ 3(ab

2c + abc

2 + a

2bc) = 3(a + b + c) ∙ abc.


247

So we have ,)(32)(2 222222 abccbacbacabcabcba

That is .)(32)( 2222 abccbacbacba

Note that a + b + c = 1, we are done.


Since )(2)(222222222 babababaabba

That is, 2

)(2

22 baba

.

Taking the square roots in both sides: )(2

2

2

222 bababa (1)

Similarly we get: )(2

222 cbcb (2)

)(2

222 acac (3)

Adding (1), (2), and (3):

)(2222222 cbaaccbba


Method 1:

)1)(1()1

1)(1

1(y

yx

x

yx

yx

)(25)2)(2(y

x

x

y

y

x

x

y

9225

Method 2:

We know that4

11 xyyx .

.91

11

1

.9812

11

1111

11

11

1

yx

xyxyxy

yx

xyyxyx



248

)3

(3)2

(2 3 abccba

abba

332 abccab

332 abcabc .

We know that 33 33 abcababcababc is true. Therefore, the original

inequality is true.


3 cba 3)( 2 cba or： 2222 acbcab

∵ baab 2 cbbc 2 caac 2 or

2)()()(222 cacbbaacbcab .

Therefore the original inequality is true.


Since a b > 0, b c > 0, a c > 0, we assume that a b = x, b c = y (x, y > 0).

Thus a c= x + y. The original inequality becomes: yxyx

411

4)11

)(( yx

yx 42 x

y

y

x.

We know that 2x

y

y

x. Therefore the original inequality is true.

Equality occurs when x = y.


We assume that (1－a)b > 4

1， (1－b)c >

4

1，and（1－c)a >

4

1.

Since a is less than 1, 1－a＞0.

2

1

4

1)1(

2

)1(

ba

ba 1－a + b > 1 (1)

Similarly, we get 1 b + c > 1 (2)

1 c + a > 1 (3)

Adding (1), (2), and (3) we get: 3 > 3 that is not possible.

Therefore not all of (1－a)b, (1－b)c, and (1－c)a greater than 1/4.

50 amc lectures problems book 1 (25) am-gm ...50 amc lectures problems book 1 (25) am-gm...

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