Page 111 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

4
s
t
2
2
2
3.70 Let r,s,t satisfy 1 ≤ r ≤ s ≤ t ≤ 4 , find the minimum value of (r − 1) +( − 1) +( − 1) +( − 1) 2
t s r
s
4
t
(r − 1) +( − 1) +( − 1) +( − 1) .
2
2
2
2
t s r
t
s
4
t
s
4
s
2
2
2
2
2
2
Solution: (r−1) +( −1) +( −1) +( −1) ≥ [ (r−1)+( −1)+( −1)+( −1) ] ⇒ 4[(r−1) +( −
s
t
r
t
4
s
(r−1)+( −1)+( −1)+( −1)
2
t 2
t
s
s
s 2
4
r2
2
(r−1) 2 t +( −1) +( −1) +( −1) ≥ [ t t 2 s 2 4 r ] ⇒ 4[(r s −1) +( − t 2
s
2
t
2
4
2
2
4
t
r
t
s
1) +( −1) +( −1) ] ≥ [(r −1)+( −1)+( −1)+( −1)] = [(r + + + )−4]
1) +( −1) +( −1) ] ≥ [(r −1)+( −1)+( −1)+( −1)] = [(r + + + )−4] .
2
2
s 2
r
4
t
s 4
r 2
s
s
t
4
s
2
t
r
t t
s r t s r t s r
√
t
s
s
t
4
Cauchy’s Inequality implies that r + + + 4 ≥ 4 4 r · · · 4 =4 4, thus
t s r t s r
√ √
2 2
2 2
2 2
2 2
4 4
2 2
2 2
√√
s s
2 2
4 4
4[( 4[(r − 1)4[(r − 1) s s t t +( − 1) +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) ++( − 22 t t s s +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) +

22
22
22
t t
ss
22
ss
44
4[(r − 1) +( − 1) +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) +r − 1) +( − 1) +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) + 22
2
2
4 4
√ √1)
r r
t t
t t
4 4
t t
√√
2 2
2 2
2 2
( ( − 1) +( − 1) ≥ 4( 2 − 1)− 1) +( − 1) ≥ 4( 2 − 1) ss 22 rr t t
22
t t
s s ( − 1) +( − 1) ≥ 4( 2 − 1)− 1) +( − 1) ≥ 4( 2 − 1) . The equal sign is obtained if and only if
22
44
(
ss r r rr
√ √
r = 2,s =2,t =2 2. Hence, the minimum value of
√
(r − 1) +( − 1) +( − 1) +( − 1) is 4( 2 − 1) .
2
s
4
t
2
2
2
2
t s r
3.71 Real numbers a 1 ,a 2 satisfy a + a ≤ 1, show that for any real numbers b 1 ,b 2,
2
2
2
1
(a 1 b 1 + a 2 b 2 − 1) ≥ (a + a − 1)(b + b − 1) always holds.
2
2
2
2
2
2
1
2
1
Proof: If b + b − 1 > 0, since a + a ≤ 1, we have (a + a − 1)(b + b − 1) ≤ 0, then
2
2
2
2
2
2
2
2
2
1
2
1
1
1
2
2
obviously (a 1 b 1 + a 2 b 2 − 1) ≥ (a + a − 1)(b + b − 1) . If b + b − 1 ≤ 0, Mean Inequality
2
2
2
2
2
2
2
1
2
2
1
1
2
2
2
2
2
2
2
a +b 2 a +b 2 1 2 2 2 2 (1−a −a )+(1−b −b )
2
2
1 2
2
1 2

implies that a 1 b 1 ≤ 1 1 ,a 2 b 2 ≤ 2 2 . Thus a 1 b 1 + a 2 b 2 ≤ (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥ (1−a −a )+(1−b −b ) 2 ⇒
2
2
2
1
2
1
1
2
2
2 2 a 1 b 1 + a 2 b 2 ≤ 1 2 (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥ (1−a −a )+(1−b −b ) 2 ⇒
2
2 2
1 2
2
1 2
2
2
2
2
2 2
1
2 1
2 2
1
2
a 1 b 1 + a 2 b 2 ≤ 2 2 (a + a + b + b ) ≤ 1 ⇒ 1 1− a 1 b 1 − a 2 b 2 ≥ 1 2 2 2 2 2 2 ⇒ 2
2 2
(1−a −a )+(1−b −b ) 2
2
] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
2 2
2
1
2 2
2
1 2 1
2
2
1 2
2
2
2
2
2
2
(1−a −a )+(1−b −b ) )
2
2
2 2
2
2
2 2
2
2
1
1
1 1 2 2 2 2 2 2 2 2 (1−a −a )+(1−b −b 2 2 2 2 (1 − a 1 b 1 − a 2 b 2 ) ≥ [ (1−a −a )+(1−b −b ) 2 1 2 [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
1
2
⇒
2
1
2
2
1
1 2
1
2
1 2
2 2
1
a 1 b 1 + a 2 b 2 ≤ (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥
a 1 b 1 + a 2 b 2 ≤ (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥ ⇒ (1 − a 1 b 1 − a 2 b 2 ) ≥ [ (1−a −a )+(1−b −b ) 2 ] ≥ 1 1 2 2 2 2 2 1 2 2 2 2 1
2
2
2
2
2
1
1
2
2
a − 1)(b + b − 1)
2 2 1 1 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 (1 − a 1 b 1 − a 2 b 2 2 ) ≥ [ 2 2 ] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
1
2
1
2
2
2
1
2
2
2
a − 1)(b + b − 1)
1
2
2
2
2
2
1
1
(1−a −a )+(1−b −b ) 2
2
2
2 2
2 2
2
2 2
2 2
(1 − a 1 b 1 − a 2 b 2 ) ≥ [ [
] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
2
1
2
2
2
1
1
1
2
2
1
2
2
(1 − a 1 b 1 − a 2 b 2 ) ≥ (1−a −a )+(1−b −b ) 2 2 ] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a + a − 1)(b + b − 1) .

2
1
1
2
1
1
2
2
1
1
2
2
2
2
2
2
1
2
2
2
2
2
2
a − 1)(b + b − 1)
a − 1)(b + b − 1)
1
2
2
2
2
1
1 a log 2
1 x(x−a+1)
3.72 Let A = {x|1+ 1 − 1 < 0}; B = {x|( ) 3 < ( ) ,a ∈ R} ,
log x log x 3 2
3
5
find the range of a such that A ⊆ B.
1
3
−1
1
Solution: 1+ log 3 x − log 5 x < 0 ⇒ 1 + log 3 − 2 log 5 < 0 ⇒ log x 25 < log x . Thus x> 1,
x
x
x
1
1 x(x−a+1)
1 a log 2
then > 3 , then 1 <x< 25 . Hence, A = {x|1 < x< 25 }. ( ) 3 < ( ) log 2 −a < 2 −x(x−a+1) ⇒ 2 −a < 2 −x(x−a+1) ⇒−a<
3
x 25 3 3 1 a log 2 1 x(x−a+1) ⇒ 3 −a −x(x−a+1) −a −x(x−a+1)
3
2
log 2
( ) 3 < ( ) ⇒ 3 3 < 2 ⇒ 2 < 2 ⇒−a<
−x(x − a + 1) ⇒ (x − a)(x + 1) < 0
2
3

1 a log 2
1 x(x−a+1)

( ) 3 < ( ) ⇒ 3 log 2 −a < 2 −x(x−a+1) ⇒ 2 −a < 2 −x(x−a+1) ⇒−a< −x(x − a + 1) ⇒ (x − a)(x + 1) < 0 ( ).
3
3 2
−x(x − a + 1) ⇒ (x − a)(x + 1) < 0
When a = −1, ( ) has no solution.
When a> −1, ( ) has the solution −1 < x <a .
When a< −1, ( ) has the solution a<x< −1.
⎧
⎨ φ (a = −1)
25
Hence, B = {x|− 1 <x <a} (a> −1) from which we know that when a ≥ 3 , A ⊆ B .
{x|a< x< −1} (a< −1)
⎩
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