DYSM 1

Problem 1Let a, b, c be positive real numbers. Prove that :

a3 + b3 + c3 + 6ab + bc + ca

(b2

a+

c2

b+

a2

c

)≥ 9

Proposed by Kunihiko Chikaya

Solution by K.Non 10 point/10

By using the AM −GM inequality :

a3 + 1 + 1 ≥ 3a

b3 + 1 + 1 ≥ 3b

c3 + 1 + 1 ≥ 3c

∴ a3 + b3 + c3 + 6 ≥ 3(a + b + c) · · · [1]

and

b2

a+ a ≥ 2b

c2

b+ a ≥ 2c

a2

c+ c ≥ 2a

∴ b2

a+

c2

b+

a2

c≥ a + b + c · · · [2]

From [1], [2], (a3 + b3 + c3 + 6)(

b2

a+

c2

b+

a2

c

)≥ 3(a + b + c)2

Using well known inequality, (a + b + c)2 ≥ 3(ab + bc + ca)

(a3 + b3 + c3 + 6)(

b2

a+

c2

b+

a2

c

)≥ 9(ab + bc + ca)

∴ a3 + b3 + c3 + 6ab + bc + ca

(b2

a+

c2

b+

a2

c

)≥ 9. Q.E.D.

Solution by Slobodan Filipovski 10 point/10

By the AM-GM inequality,

a3 + b3 + c3 + 6 = (a3 + 1 + 1) + (b3 + 1 + 1) + (c3 + 1 + 1)

≥ 3 3√

a3 · 1 · 1 + 3 3√

b3 · 1 · 1 + 3 3√

c3 · 1 · 1 = 3(a + b + c).

From Cauchy-Schwarz inequality, we haveb2

a+

c2

b+

a2

c≥ (a + b + c)2

a + b + c= a+b+c.

∴ (a3 + b3 + c3 + 6)ab + bc + ca

(b2

a+

c2

b+

a2

c

)≥ 3(a + b + c)

ab + bc + ca·(a+b+c) =

3(a + b + c)2

ab + bc + ca,

Since (a− b)2 + (b− c)2 + (c− a)2 ≥ 0 ⇐⇒ a2 + b2 + c2 ≥ ab + bc + ca,thus weobtain

3(a+b+c)2 = 3(a2 +b2 +c2 +2ab+bc+2ca) ≥ 3(ab+bc+ca+2ab+2bc+2ca) =9(ab + bc + ca).

∴ (a3 + b3 + c3 + 6)ab + bc + ca

(b2

a+

c2

b+

a2

c

)≥ 9(ab + bc + ca)

ab + bc + ca≥ 9.

Equality holds when a3 = b3 = c3 = 1 ⇐⇒ a = b = c = 1.

2

Solution by Time Spring 8 point/10

One application of Cauchy’s Inequality, show that :

a3 + 2 = a3 + 1 + 1 ≥ 3 · a3︸︷︷︸miss typing

·1 · 1 = 3a

b3 + 2 = b3 + 1 + 1 ≥ 3 · b3︸︷︷︸miss typing

·1 · 1 = 3a

c3 + 2 = b3 + 1 + 1 ≥ 3 · c3︸︷︷︸miss typing

·1 · 1 = 3a

∴ a3 + b3 + c3 + 6 ≥ 3(a + b + c) · · · (1)

In addition, One application of Cauchy-Bunyakovsky-Schwarz Inequality,

show that :

[(b√a

)2

+(

c√b

)2

+(

a√c

)2]

[(√

a)2 + (√

b)2 + (√

c)2]

≥

b√a· √a +

b︸︷︷︸miss typing√

b·√

b +a√c· √c

2

⇐⇒(

b2

a+

c2

b+

a2

c

)≥ (a + b + c)2

a + b + c= a + b + c · · · (2)

Then, with (1), (2), we have

(a3 + b3 + c3 + 6)ab + bc + ca

(b2

a+

c2

b+

a2

c

)≥ 3(a + b + c)2

ab + bc + ca

≥ 3 · 3(ab + bc + ca)ab + bc + ca

= 9.

3

Solution by Proposer

a3 ≥ 3a− 2 ⇐⇒ (a− 1)2(a + 2) ≥ 0 etc, we have

b2

3a+

c2

3b+

a2

3c≥ b2

a3 + 2+

c2

b3 + 2+

a2

c3 + 2≥ (a + b + c)2

a3 + b3 + c3 + 6≥ 3(ab + bc + ca)

a3 + b3 + c3 + 6.

Since a, b, c are positive, yielding

(a3 + b3 + c3 + 6)ab + bc + ca

(b2

a+

c2

b+

a2

c

)≥ 9(ab + bc + ca).

4

Problem 3

Let θ be real number. Find the sum∞∑

n=1

4n sin4

(θ

2n

).

Solution by K.Non 8 point/10

For real number a, from sin 2a = 2 sin a cos a, we have

4 sin2 a− sin2 2a = 4 sin2 a− 4 sin2 a cos2 a = 4 sin4 a

Substitute a =θ

2nfor this equation, we obtain

4n sin2

(θ

2n

)− 4n−1 sin2

(θ

2n−1

)= 4n sin4

(θ

2n

)

,

thus, we havek∑

n=1

4n sin4

(θ

2n

)=

k∑

n=1

{4n sin2

(θ

2n

)− 4n−1 sin2

(θ

2n−1

)}

= 4k sin2

(θ

2k

)− sin2 θ.

∞∑

n=1

4n sin4

(θ

2n

)= lim

k→∞

k∑

n=1

4n sin4

(θ

2n

)= lim

k→∞4k sin2

(θ

2k

)− sin2 θ

= limk→∞

sin2 θ2k(

θ2k

)2 · θ2 − sin2 θ with θ 6= 0︸ ︷︷ ︸−1 point

Letθ

2k= c,

= limc→0

(sin c

c

)2

· θ2 − sin2 θ

∴∞∑

n=1

4n sin4

(θ

2n

)= θ2 − sin2 θ,

which satisfies the result of the case θ = 0︸ ︷︷ ︸−1 point

.

5

Solution by Kunihiko Chikaya

For k = 1, 2, · · · , Using identities, sin2 x =1− cos 2x

2, cos2 x =

1 + cos 2x2

sin4 θ

2k=

(sin2 θ

2k

)2

=

(1− cos θ

2k−1

2

)2

=14

(1− cos

θ

2k−1

)2

=14

(1− 2 cos

θ

2k−1+ cos2

θ

2k−1

)

=14

(1− 2 cos

θ

2k−1+

1 + cos θ2k−2

2

)

∴ Sn :=n∑

k=1

4k sin4 θ

2k

=n∑

k=1

4k−1

(32− 2 cos

θ

2k−1+

12

cosθ

2k−2

)

=32

n∑

k=1

4k−1 +n∑

k=1

(22k−3 cos

θ

2k−1− 22k−1 cos

θ

2k−1

)

=32· 4n − 1

4− 1

+12

cos 2θ − 2 cos θ

+2 cos θ − 23 cosθ

2

+23 cosθ

2− 25 cos

θ

22

+22n−5 cosθ

2n−3− 22n−3 cos

θ

2n−2

· · · · · ·

6

+22n−3 cosθ

2n−2− 22n−1 cos

θ

2n−1

=4n − 1

2+

12

cos 2θ − 22n−1 cosθ

2n−1

= 22n−1

(1− cos

θ

2n−1

)− 1

2(1− cos 2θ)

=1− cos θ

2n−1(θ

2n−1

)2 · 2θ2 − sin2 θ with θ 6= 0.

Since limx→0

1− cosx

x2= lim

x→0

1− cos2 x

x2(1 + cosx)= lim

x→0

(sinx

x

)2

· 11 + cosx

= 12 · 12

=12, using this gives

∞∑

n=1

4n sin4

(θ

2n

)2

= limn→∞Sn

=12· 2θ2 − sin2 θ

= θ2 − sin2 θ, which satisfies the result of the case θ = 0.

7