Inequality 1 Sol

Pure Mathematics – Inequalities 1

p.1

Inequalities Assignment 1 – Suggested Solution 1. (a) x −6 ≤3 −3≤ x −6 ≤ 3 3≤ x ≤9

(x ≥ 3 or x ≤ −3)

and (− 9 ≤ x ≤ 9 ) ∴ − 9 ≤ x ≤ −3 or 3 ≤ x ≤ 9

Case 1: x ≥6 x −6≤3 x ≤9 −9 ≤ x ≤ 9 ∴ − 9 ≤ x ≤ −6 or 6 ≤ x ≤ 9 Case 2: x <6

− ( x − 6) ≤ 3 x ≥3 x ≥ 3 or x ≤ −3 ∴ − 6 ≤ x ≤ −3 or 3 ≤ x ≤ 6 Thus, the required solution is − 9 ≤ x ≤ −3 or 3 ≤ x ≤ 9 .

(b) Putting x = 1 – 2y, we have − 9 ≤ 1 − 2 y ≤ −3 or 3 ≤ 1 − 2 y ≤ 9 − 10 ≤ −2 y ≤ −4 or 2 ≤ −2 y ≤ 8 2 ≤ y ≤ 5 or − 4 ≤ y ≤ −1 ∴ − 4 ≤ y ≤ −1 or 2 ≤ y ≤ 5 2.

(a)(i)

a + ak – (1 + ak+1) = = =

a – 1 – ak+1 + ak a – 1 – ak(a – 1) (a – 1)(1 – ak) ⎧< 0 if 0 < a < 1 ⎪ ⎨= 0 if a = 1 ⎪> 0 if a > 1 ⎩

≤0 Thus, a + a ≤ 1 + a k +1 . k

Pure Mathematics – Inequalities 1

(

a + a k − 1 + a k +1

p.2

)

= −a k +1 + a k + a − 1 = −a k (a − 1) + a − 1

(

) = −(a − 1)(a − 1)(a = −(a − 1) (a + a = −(a − 1) a k − 1

k −1

2

k −1

)

+ a k − 2 + a k −3 + L + 1

k −2

≤0

)

+ a k −3 + L + 1

Thus, a + a k ≤ 1 + a k +1 . (ii) ∵ 1 + a = 1 + a ∴ The statement is true for n = 1.

(

Assume that (1 + a ) ≤ 2 k −1 1 + a k k

)

for some positive integer k.

Then, we have

(1 + a )k +1

( ) (1 + a + a (1 + 1 + a (1 + a )

≤ 2 k −1 1 + a k (1 + a ) = 2 k −1

k

≤ 2 k −1

k +1

= 2k

(by induction assumption )

+ a k +1

)

+ a k +1

) (by(a)(i))

k +1

(

Thus, by induction, (1 + a ) ≤ 2 n −1 1 + a n n

(b) Note that

)

for any positive integer n.

y y in (a)(ii), we have > 0 . Putting a = x x

n ⎛ ⎛ y ⎞n ⎞ y⎞ ⎛ n −1 ⎜ ⎟ 1 2 1+ ≤ + ⎟ ⎜ ⎜ ⎜⎝ x ⎟⎠ ⎟ x⎠ ⎝ ⎝ ⎠

(x + y )n ≤ 2 n−1 (x n + y n )

xn + yn ⎛x+ y⎞ ⎜ ⎟ ≤ 2 ⎝ 2 ⎠ n

3.

(a) Note that tn – 1 = (t – 1)(tn-1 + tn-2 + … + t + 1). Case 1: 0 < t < 1 Under this case, we have t – 1 < 0 and 0 < tn-1 + tn-2 + … + t + 1< n. So we have (t – 1)( tn-1 + tn-2 + … + t + 1) > n(t – 1). Therefore, we have tn – 1 > n(t – 1). Case 2: t = 1 Under this case, we have tn – 1 = n(t – 1).

Pure Mathematics – Inequalities 1

p.3

Case 3: t > 1 Under this case, we have t – 1 > 0 and tn-1 + tn-2 + … + t + 1 > n. So we have (t – 1)( tn-1 + tn-2 + … + t + 1) > n(t – 1). Therefore, we have tn – 1 > n(t – 1). Thus, by combining the three cases, we have tn – 1 > n(t – 1) for all t > 0. Since t – 1 = t – 1, the result is true when n = 1. When n ≥ 2, let f(t) = tn – nt for all t > 0. Then, we have f’(t) = ntn-1 – n = n(tn-1 – 1) ⎧< 0 if ⎪ ⎨= 0 if ⎪> 0 if ⎩

0 < t <1 t =1 t >1

So, f(t) attains its greatest value when t = 1. Therefore, f(t) ≥ f(1) for all t > 0 tn – nt ≥ 1 – n for all t > 0 tn – 1 ≥ n(t – 1) for all t > 0 (b) (i) By (a) 3

⎛ 3 abc ⎞ ⎛3 ⎜ ⎟ − 1 ≥ 3⎜ ⎜ ab ⎟ ⎜ ⎝ ⎠ ⎝ ⎛3 c − 1 ≥ 3⎜⎜ ab ⎝

⎞ − 1⎟⎟ ab ⎠ abc ⎞ ⎟−3 ab ⎟⎠ abc

c ≥ 33 abc − 2 ab ⎛a+b+c⎞ ⎛a+b⎞ 3 3⎜ ⎟ ≥ 3 abc − 2 ab ⎟ − 2⎜ 3 ⎠ ⎝ 2 ⎠ ⎝ ⎛a+b+c⎞ 3 ⎛a+b⎞ 3⎜ ⎟ − 3 abc ≥ 2⎜ ⎟ − 2 ab 3 ⎝ ⎠ ⎝ 2 ⎠ a+b+c 3 2⎛a+b ⎞ − abc ≥ ⎜ − ab ⎟ 3 3⎝ 2 ⎠ (ii) By (a), putting n = k + 1 and t = k +1

⎞ ⎛G − 1 ≥ (k + 1)⎜⎜ k +1 − 1⎟⎟ ⎠ ⎝ Gk ⎞ ⎛G y k +1 − 1 ≥ (k + 1)⎜⎜ k +1 − 1⎟⎟ Gk ⎠ ⎝ Gk

⎛ Gk +1 ⎞ ⎟⎟ ⎜⎜ G k ⎠ ⎝

y k +1 ≥ (k + 1)Gk +1 − kGk

Gk +1 , Gk

Pure Mathematics – Inequalities 1

p.4

(iii) Since x1 = x1, the statement is true when n = 1. Assume that there exists a positive integer k such that x1 + x 2 + L + x k k ≥ x1 x 2 L x k for any k positive real numbers x1, k

x2, …, xk. Then, for any k + 1 positive real numbers y1, y2, …, yk+1, define Al =

y1 + y 2 + L + y l and Gl = l y1 y 2 L y l for l = k, k + 1. l

Note that yk+1 = (k + 1)Ak+1 – kAk. By (b)(ii), we have

(k + 1)Ak +1 − kAk ≥ (k + 1)Gk +1 − kGk (k + 1)( Ak +1 − Gk +1 ) ≥ k ( Ak − Gk ) Ak +1 − Gk +1 ≥

k ( Ak − Gk ) k +1

By putting xi = yi (1 ≤ i ≤ k) in the induction assumption, we have Ak ≥ Gk. Therefore, we have Ak+1 ≥ Gk+1 if Ak ≥ Gk. So the statements is true for n = k + 1 if the statement is true for n = k. Thus, by mathematical induction, the statement is true for all positive integers n. 4.

(a) Since a ≥ 0, b ≥ 0, n ≥ 2

(a + b )n

= a n + C1n a n −1b + C 2n a n − 2 b 2 + L + C nn b n ≥ a n + na n −1b

The equality holds if and only if C 2n a n − 2 b 2 = C 3n a n −3b 3 = L = C nn b n = 0 if and only if b = 0. (b) (i) Ak +1 − Ak a1 + a 2 + a3 + L + a k +1 a1 + a 2 + a3 + L + a k − k +1 k ka k +1 − (a1 + a 2 + a3 + L + a k ) = k (k + 1) (a − a1 ) + (a k +1 − a 2 ) + (a k +1 − a3 ) + L + (a k +1 − a k ) = k +1 ≥0 k (k + 1) =

Thus, Ak+1 ≥ Ak for all k = 1, 2, 3, ….

Pure Mathematics – Inequalities 1

p.5

Ak +1 = = ≥ ≥ =

kAk + a k +1 k +1 a − Ak Ak + k +1 k +1 a − ak (Q Ak ≤ a k ) Ak + k +1 k +1 a − a k +1 (Q a k ≤ a k +1 ) Ak + k +1 k +1 Ak for all k = 1, 2, 3, K.

(ii) By (b)(i), Ak+1 - Ak ≥ 0 Note that Ak > 0 and Gk > 0 for all k ∈ N.

Ak +1

k +1

= ( Ak + ( Ak +1 − Ak ))

k +1

≥ Ak

k +1

+ (k + 1)Ak ( Ak +1 − Ak ) k

= Ak ( Ak + (k + 1)( Ak +1 − Ak )) k

= Ak

k

((k + 1)Ak +1 − kAk )

= Ak ak +1 k

Thus, Ak+1k+1 ≥ Akkak+1 for all k = 1, 2, 3, …. An = Gn if and only if a1 = a2 = a3 = … = an for all n = 1, 2, 3, …. When n = 1, A1 = a1 = G1. Therefore, the latter result is true for n = 1. Assume that Ak ≥ Gk and Ak ≥ Gk if and only if a1 = a2 = a3 = … = ak, where k is a positive integer. Then, we have Ak +1

k +1

≥ Ak a k +1 k

= Gk a k +1 k

= a1 a 2 a 3 L a k +1 = Gk +1

k +1

Therefore, Ak+1 ≥ Gk+1. Moreover, if a1 = a2 = a3 = … = ak+1, then a1 + a 2 + L + a k +1 = a1 and Gk +1 = k +1 a1 a 2 L a k +1 = a1 k +1 Ak+1 = Gk+1.

Ak +1 =

∴

Pure Mathematics – Inequalities 1

p.6

Furthermore, if Ak+1 = Gk+1, then

( Ak + ( Ak +1 − Ak ))k +1 = Ak k +1 + (k + 1)Ak k ( Ak +1 − Ak ) ∴

and Ak ≥ Gk.

Ak+1 – Ak = 0 (by (a)) and a1 = a2 = a3 = … = ak (by induction assumption)

∴ ∴ ∴

a1 + a 2 + L + a k +1 a1 + a 2 + L + a k = and a1 = a2 = a3 = … = ak k +1 k a1 = a2 = a3 = … = ak = ak+1

Ak+1 = Gk+1 if and only if a1 = a2 = a3 = … = ak+1.

Thus, by mathematical induction, the latter result is true for all n = 1, 2, 3, …. (c) Put a1 = 1, a2 = a3 = a4 = … = an = an+1 = Since a1 < a2 = a3 = a4 = … = an = an+1 ⎛ n + 1⎞ 1 1 + n⎜ ⎟ ⎛ n n +1 ⎝ n ⎠ > ⎜ (1)⎛ n + 1 ⎞ ⎞⎟ ⎜ ⎜⎝ n ⎟⎠ ⎟ n +1 ⎝ ⎠ n

n + 2 ⎛ n + 1 ⎞ n +1 >⎜ ⎟ n +1 ⎝ n ⎠ n

1 ⎛ 1 ⎞ n +1 > ⎜1 + ⎟ Therefore, 1 + n +1 ⎝ n ⎠

1 ⎞ ⎛ Thus, ⎜1 + ⎟ ⎝ n +1⎠

n +1

n

⎛ 1⎞ > ⎜1 + ⎟ . ⎝ n⎠

n +1 . n