Test 6 - Sol

Pure Mathematics – Inequalities

p.1

King’s College 2007 – 2008 F.6 Pure Mathematics Revision Test 6 Suggested Solution n

1.

(a) (i)

n

n

n

k =1

k =1

Consider f (t ) = ∑ (a k + tbk ) = ∑ a k + 2t ∑ a k bk + t 2 ∑ bk 2

k =1

2

k =1

2

Q f (t ) ≥ 0 ∀t ∴ Δ≤0 2

⎧ n ⎫ ⎧ n 2 ⎫⎧ n 2 ⎫ ⇔ ⎨2∑ a k bk ⎬ − 4⎨∑ a k ⎬⎨∑ bk ⎬ ≤ 0 ⎩ k =1 ⎭ ⎩ k =1 ⎭⎩ k =1 ⎭ 2

⎧n ⎫ ⎧ n 2 ⎫⎧ n 2 ⎫ ⇔ ⎨∑ a k bk ⎬ ≤ ⎨∑ a k ⎬⎨∑ bk ⎬ ⎩ k =1 ⎭ ⎩ k =1 ⎭⎩ k =1 ⎭

(ii)

p≤

bk ≤ q , k = 1, 2, …, n ak

⎛ b ⎞⎛ b ⎞ ⇒ a k2 ⎜⎜ p − k ⎟⎟⎜⎜ q − k ⎟⎟ ≤ 0 k = 1, 2, K , n a k ⎠⎝ ak ⎠ ⎝ ⇒ pqa k2 − ( p + q )a k bk + bk2 ≤ 0 ⇒

∑ [pqa − ( p + q )a b n

2 k

k =1

⇒

k

n

(iii)

]

+ bk2 ≤ 0

n

n

pq ∑ a k2 − ( p + q )∑ a k bk + ∑ bk2 ≤ 0 k =1

⇒

k

k =1

k =1

n

n

n

k =1

k =1

k =1

pq ∑ a k2 + ∑ bk2 ≤ ( p + q )∑ a k bk

m bk M ≤ ≤ M ak m

By (b)(ii), n n ⎛m M⎞ n 2 2 ⎜ + ⎟∑ a k bk ≥ ∑ a k + ∑ a k M m ⎝ ⎠ k =1 k =1 k =1

⎛ n ⎞⎛ n ⎞ ≥ 2 ⎜ ∑ a k2 ⎟⎜ ∑ bk2 ⎟ ⎝ k =1 ⎠⎝ k =1 ⎠ 2

2

1⎛ m M ⎞ ⎛ n ⎞ ⎛ n 2 ⎞⎛ n 2 ⎞ ⎜ + ⎟ ⎜ ∑ a k bk ⎟ ≥ ⎜ ∑ a k ⎟⎜ ∑ bk ⎟ 4 ⎝ M m ⎠ ⎝ k =1 ⎠ ⎝ k =1 ⎠⎝ k =1 ⎠

(Q A.M . ≥ G.M .)

Pure Mathematics – Inequalities

(b) Choose a k = 1 +

p.2

1 8 1 4 1 1 , bk = 1 − k +1 , then 1 − 2 = ≤ a k , bk ≤ 1 + = . k 9 3 3 3 3 3

8 4 Take m = , M = . 9 3 By (a)(iii)

By (a)(i),

⎧⎪ n ⎛ 1 ⎞ 1 ⎞ ⎪⎫⎪⎧ n ⎛ 1 − k +1 ⎟ 1 + ⎜ ⎨∑ k ⎟ ⎬ ⎨∑ ⎜ ⎪⎩ k =1 ⎝ 3 ⎠ ⎪⎭⎪⎩ k =1 ⎝ 3 ⎠ 2

2

⎫⎪ ⎬ ⎪⎭

2 2 ⎧⎪ n ⎛ 1 ⎞ ⎫⎪⎧⎪ n ⎛ 1 ⎞ ⎫⎪ ⎨∑ ⎜1 + k ⎟ ⎬⎨∑ ⎜1 − k +1 ⎟ ⎬ ⎪⎩ k =1 ⎝ 3 ⎠ ⎪⎭⎪⎩ k =1 ⎝ 3 ⎠ ⎪⎭

2

⎛4 8⎞ 2 1 ⎜⎜ 3 9 ⎟⎟ ⎧ n ⎛ 1 ⎞⎫ 1 ⎞⎛ ≤ + ⎟⎬ ⎨∑ ⎜1 + ⎟⎜1 − 4 ⎜ 8 4 ⎟ ⎩ k =1 ⎝ 3 k ⎠⎝ 3 k +1 ⎠⎭ ⎜ ⎟ ⎝9 3⎠ 169 ⎧ n ⎛ 1 1 1 ⎞⎫ = ⎨∑ ⎜1 + k − k +1 − 2 k +1 ⎟⎬ 144 ⎩ n =1 ⎝ 3 3 3 ⎠⎭ 169 ⎧ n ⎛ 2 1 1 ⎞⎫ = ⎨∑ ⎜1 + ⋅ k − 2 k +1 ⎟⎬ 144 ⎩ n =1 ⎝ 3 3 3 ⎠⎭ 169 ⎧ 2 n 1 1 n 1⎫ = + n ⎨ ∑ − ∑ ⎬ 144 ⎩ 3 n =1 3 k 3 n =1 9 k ⎭ 169 ⎛ 2 n 1 ⎞ < ⎜n + ∑ k ⎟ 144 ⎝ 3 n =1 3 ⎠

2

1 ⎛ 1− n ⎜ 169 ⎜ 2 1 3 = n+ ⋅ ⋅ 1 144 ⎜ 3 3 1− ⎜ 3 ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

169 ⎡ 1 ⎛ 1 = n + ⋅ ⎜1 − n ⎢ 144 ⎣ 3 ⎝ 3

2

169 ⎛ 1⎞ < ⎜n + ⎟ 144 ⎝ 3⎠

⎞⎤ ⎟⎥ ⎠⎦

2

2

2

2

⎧n ⎛ 1 ⎞⎛ 1 ⎞⎫ ≥ ⎨∑ ⎜1 + k ⎟⎜1 − k +1 ⎟⎬ ⎩ k =1 ⎝ 3 ⎠⎝ 3 ⎠⎭

2

⎧n ⎛ 1 1 1 ⎞⎫ = ⎨∑ ⎜1 + k − k +1 − 2 k +1 ⎟⎬ 3 3 ⎠⎭ ⎩ n =1 ⎝ 3 2 n 1 1 n 1⎫ ⎧ = ⎨n + ∑ k − ∑ k ⎬ 3 n =1 3 3 n =1 9 ⎭ ⎩

2

2 n 1 1 n 1⎫ ⎧ > ⎨n + ∑ k − ∑ k ⎬ 3 n =1 3 3 n =1 3 ⎭ ⎩

2

1 n 1⎫ ⎧ = ⎨n + ∑ k ⎬ 3 n =1 3 ⎭ ⎩ 1 1⎞ ⎛ > ⎜n + ⋅ ⎟ 3 3⎠ ⎝ 1⎞ ⎛ = ⎜n + ⎟ 9⎠ ⎝

2

2

2

2

2 2 2 2 ⎧⎪ n ⎛ 1⎞ 1 ⎞ ⎫⎪⎧⎪ n ⎛ 1 ⎞ ⎫⎪ 169 ⎛ 1⎞ ⎛ + Therefore, ⎜ n + ⎟ < ⎨∑ ⎜1 + k ⎟ ⎬⎨∑ ⎜1 − k +1 ⎟ ⎬ < n ⎜ ⎟ 9⎠ 3⎠ ⎪⎩ k =1 ⎝ 3 ⎠ ⎪⎭⎪⎩ k =1 ⎝ 3 ⎠ ⎪⎭ 144 ⎝ ⎝

2

Pure Mathematics – Inequalities

9.

p.3

y2 y D1 , 1 D2 > 0 y1 y2

(a) (i)

By A.M . ≥ G.M . , y2 y y y D1 + 1 D2 ≥ 2 2 D1 ⋅ 1 D2 = 2 D1 D2 y2 y1 y2 y1

(ii)

y2 y D1 + 1 D2 y1 y2

(

)

(

y2 y x1 y1 − z12 + 1 x 2 y 2 − z 22 y1 y2

=

= x1 y 2 −

)

y2 2 y ⋅ z1 + x 2 y1 − 1 ⋅ z 22 y1 y2

⎛y ⎞⎛ y ⎞ ≤ x1 y 2 + x 2 y1 − 2 ⎜⎜ 2 ⋅ z12 ⎟⎟⎜⎜ 1 ⋅ z 22 ⎟⎟ ⎝ y1 ⎠⎝ y 2 ⎠ = x1 y 2 + x 2 y1 − 2 z1 z 2 (iii) (x1 + x 2 )( y1 + y 2 ) − ( z1 + z 2 )

2

= x1 y1 + x 2 y 2 + x1 y 2 + x 2 y1 − z12 − 2 z1 z 2 − z 22 = ( x1 y 2 + x 2 y1 − 2 z1 z 2 ) + (D1 + D2 )

y2 y D1 + 1 D2 + (D1 + D2 ) by (a )(ii ) y1 y2

≥

≥ 2 D1 D2 + (D1 + D2 ) by (a )(i ) ≥ 2 D1 D2 + 2 D1 D2 = 4 D1 D2 (b)

8

(x1 + x2 )( y1 + y 2 ) − (z1 + z 2 )2 ≤

8 4 D1 D2

=2

1 1 ⋅ D1 D2

≤

1 1 + D1 D2

=

1 x1 y1 − z1

2

+

1 x2 y 2 − z 2

2

by A.M . ≥ G.M .

Pure Mathematics – Inequalities

If the equality holds, then ⎧ ⎪ L (1) ⎪ D1 = D2 ⎪ y2 y1 D2 L (2) ⎨ D1 = y y 1 2 ⎪ ⎪ y 2 2 y1 2 z2 L (3) ⎪ z1 = y2 ⎩ y1 Put (1) into (2), we have

y2 y = 1 y1 y 2 ⇒ y12 = y 22 ⇒ y1 = y 2

(Q y1 , y 2 > 0)

From (3), z1 = z2 Sub into (1), x1 = x2

p.4