Page 108 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

The solution of the inequality x + ax + b< 0 should have the form α <x < β . Since the
2
solution of the inequality system is 4 ≤ x< 5, then β =5, −2 ≤ α< 4. Since β =5, then

25 + 5a + b =0 Since α + β = −a , then 3 ≤−a< 9, that is, −9 <a ≤−3. As a conclusion,

a, b should satisfy −9 <a ≤−3
5a + b + 25 = 0.

2
2
2
2
3.63 If x,y,z ≥ 1, show (x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2
(x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2.
2
2
2
2
2 2 2 2 2 2
Proof: Since x ≥ 1,y ≥ 1, we have (x −2x+2)(y −2y+2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y −

2
2
2
2
2
2
(x −2x+2)(y −2y 2 +2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y −
2
2
4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=
2
2
2
2
2
2
2
2 2y+2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y −
2
(x −2x+2)(y −
2 2 2 2 2 2 4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=
(x −2x+2)(y −2y+2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y − −2(y − 1)(x − 1)(x + y − 1) ≤ 0 2 2
2
2
2
2
2
2
2
2 −2y+2)−[(xy) −2xy+2] =
(x −2x+2)(y
−2(y − 1)(x − 1)(x + y − 1) ≤ 0
2
2(−2y+2)x +(6y−2y −4)x+(2y −
4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]= 4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=

2
2
2
2
4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=
−2(y − 1)(x − 1)(x + y − 1) ≤ 0 2 2 −2(y − 1)(x − 1)(x + y − 1) ≤ 0, then (x − 2x + 2)(y − 2y + 2) ≤ (xy) − 2xy +2
−2(y − 1)(x − 1)(x + y − 1) ≤ 0

2
2
2
2
2
2
(x − 2x + 2)(y − 2y + 2) ≤ (xy) − 2xy +2(i). Similarly, since xy ≥ 1,z ≥ 1, we have [(xy) − 2xy + 2](z − 2z + 2) ≤ (xyz) − 2xyz +2
2
2
2
2
2
2
2
[(xy) − 2xy + 2](z − 2z + 2) ≤ (xyz) − 2xyz +2 (ii). From (i) and (ii), we can obtain (x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2

(x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2.
2
2
2
2
1
1
3.64 Given natural numbers a<b< c , m is an integer, and + + 1 = m , find
a b c
a, b, c .
1 1 1 5
Solution: Since a, b, c are natural numbers and a<b< c , we have a ≥ 1,b ≥ 2,c ≥ 3, 0 <m ≤ + + =1
1 2 3 6
1 1 1 5
a ≥ 1,b ≥ 2,c ≥ 3, 0 <m ≤ + + =1 . Since m is an integer, we have m =1 and a =1 . If a ≥ 3 , then
1 2 3 6
1
1
1
1 + + 1 1 + + 1 = 47 < 1. Hence, + + 1 = m =1. Therefore, a =2. Then + 1 1 = 1
1
1
a b c ≤ 3 4 5 60 a b c b c =1 − 2 2
1 + 1 1 = . If b ≥ 4, then + 1 1 + 1 = 9 1 1 1 1 = ,
1
1
1
b c =1 − 2 2 b c ≤ 4 5 20 < , thus b =3. Then =1 − − 3 6
c
2
2
thus c =6.
x
x
3.65 Given 1 <a < 2,x ≥ 1, and f(x) = a +a −x ,g(x)= 2 +2 −x . (1) Compare f(x)
2 2
m .
n 1 1
and g(x). (2) Let n ∈ N, n ≥ 1, show f(1) + f(2) + ··· + f(2n) < 4 − 2n
2 2
x
x
x x
x
x 2x
x
x
a −a
+1
1 a
Solution: (1)f(x)−g(x) = a +a −x − 2 +2 −x = ( 2x +1 2 2x x ) = 2 a +2 −2 2x x x = (a −2 )(2 a −1) .
2 2 2 a x − 2 2 x+1 x 2 x+1 x
a
a
x
x x
Since 1 <a < 2,x ≥ 1, then 2 a > 1,a < 2 , thus (a −2 )(2 a −1) < 0 , that is,
x
x x
x
x
a
2 x+1 x
f(x) − g(x) < 0. Hence, f(x) <g(x).
2n
1 1
1
2
(2) Since f(x) <g(x), then f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= (2 + 2 + ···+2 )+ ( +
1 1
1
2n
2
f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= 2 2 (2 + 2 + ···+2 )+ ( 2 2+
1
n 2 2
1
1
1
1
1
2n−1
n
2
2 + ··· +
2n ) = (1 +2 +2 + ··· +2
1
2n =4 − 1
2
n
)+ (1 − 1
1
1 1
2
2n ) < 4 − 1 +1 − 1
n
2n
f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= (2 + 2 + ···+2 )+ ( + 1 2 + 2 ··· + 1 2n ) 2 = (1 +2 +2 + ··· +2 2n−1 )+ (1 2 − 2 ) < 4 − 1 +1 − 2 =4 − 2 2n
1
1 1
2n
2
f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= 2 (2 + 2 + ···+2 )+ ( + 2 2 2 2 2n 2 2n 2 2n
2 2
1 1 2 2n−1 1 1 2 n 1 n 2 2 1
2 + ··· + 1
)+ (1 − 1
2n ) = (1 +2 +2 + ··· +2 2n−1
2
n
n
1 2 + ··· + 2 ) = (1 +2 +2 + ··· +2 )+ 1 2n ) < 4 − 1 +1 − 1 2n =4 − 1 2 .
2n
2 =4 −
2 ) < 4 − 1 +1 −
2 2 2 2n 2 2 (1 − 2 2n 2 2n 2 2n
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