Page 107 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.58 a 1 ,a 2 , ··· ,a n are positive numbers and satisfy a 1 a 2 ··· a n =1 , show
(2 + a 1 )(2 + a 2 ) ··· (2 + a n ) ≥ 3 .
n
√
3
Proof: Use an arithmetic mean-geometric mean inequality a + b + c ≥ 3 abc (a, b, c are
√
positive numbers) to obtain 2+ a i = 1+ 1+ a i ≥ 3 a i (i =1, 2, ··· ,n ). Then
3
√√
n n 3 3 n n

(2 + a 1 )(2 + a 2 ) ··· (2 + a n ) ≥ 3 ·(2 + a 1 )(2 + a 2 ) ··· (2 + a n ) ≥ 3 · a 1 a 2 ··· a n =3a 1 a 2 ··· a n =3 .
3.59 If a, b, c are side lengths of a triangle, show a b(a − b)+ b c(b − c)+ c a(c − a) ≥ 0
2
2
2
and determine when the equal sign is reached.
Proof: Let a = y + z, b = z + x, c = x + y where x,y,z are positive numbers. Substitute

them into the inequality:
22 22 22
(y + z) (x + z)(y − x)+(z + x) (x + y)(z − y)+ (x + y) (y + z)(x − z) ≥ 0 ⇔(y + z) (x + z)(y − x)+(z + x) (x + y)(z − y)+ (x + y) (y + z)(x − z) ≥ 0 ⇔
2
2
2
(y + z) (x + z)(y − x)+(z + x) (x + y)(z − y)+ (x + y) (y + z)(x
− z) ≥ 0 ⇔
33
33
33
x
x z + y x + z y − xyz(x + y + z) ≥ 0z + y x + z y − xyz(x + y + z) ≥ 0
x z + y x + z y − xyz(x + y + z) ≥ 0. Divide both sides by xyz to obtain
3
3
3
x 2 + y 2 + z 2 ≥ x + y + z which can be proven by the inequalities
y z x
2 2
2 2
2 2
xx + y ≥ 2x,+ y ≥ 2x, yy + z ≥ 2y,+ z ≥ 2y, zz + x ≥ 2z+ x ≥ 2z .
yy zz xx
These inequalities have the equal sign if and only if x = y = z , that is, the original inequality
has the equal sign if and only if a = b = c .
|a+b| |a| + |b|
≤ 1+|b| .
3.60 a, b are real numbers, show 1+|a+b| 1+|a|
=
=
|a+b| 1+|a+b|−1 1 1 |a|+|b| |a| |b|
+
1
1
|a|
|a|+|b|
|b|
1+|a+b| =
Proof: Since |a + b| ≤|a| + |b|, we have |a+b| = 1+|a+b|−1 =1 − 1+|a+b| ≤ 1 − 1+|a|+|b| = 1+|a|+|b| = 1+|a|+|b| + 1+|a|+|b| ≤
≤
≤ 1 −
1+|a+b| =1 −
1+|a+b| 1+|a+b| 1+|a+b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b|
|a+b| 1+|a+b|−1 1 1 |a|+|b| |a| |b| |a| + |b|
= =1 − ≤ 1 − = = + ≤ 1+|a| + 1+|b|.
|a|
|b|
1+|a+b| 1+|a+b| 1+|a+b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b| 1+|a| 1+|b|
|a| |b|
+
1+|a| 1+|b|
1
y
x
2
3.61 Given 0 <a < 1,x + y =0, show log (a + a ) ≤ log 2+ .
a
a
8
√ x+y
y
x+y
x
Proof: a + a ≥ 2 a a =2a 2 . Since 0 <a < 1, we have log (a + a ) ≤ log (2a x+y x+y x+y = log 2+ x−x 2 = log 2+ x(1 − x) ≤
x y
1
x
y
2 ) = log 2+
2
2
a
x
a x
y
y
2 ) = log 2+) = log 2+
log (a +(a + a ) ≤ log (2a 2 a x+y 2 x+y a x−x 2x−x a 1 2 1
= log 2+= log 2+
1 x+1−x 2 a ) ≤ log (2alog
= log 2+ x(1 − x) ≤= log 2+ x(1 − x) ≤
a
a
a
(
log a 2x+1−x 2 ) = log 2+ 1 a a a 2 2 a a 2 2 a a 2 2
x+y
) = log 2+) = log 2+ .
1
a
1
2
y
1
x
(
log
log (a + a ) ≤ log (2a 2 ) = log 2+ x+y = log 2+ x−x = log 2+ x(1 − x) ≤ log 2+ ( 2 x+1−x 2 a a 1 8 8 1

8
a
a
a
2
a 2 a 2
2
a
a
2
2
2
log 1 x+1−x 2 1
(
) = log 2+
a 2 2 a 8
3.62 The system of inequalities
√
2
x − 2x − 8 < 8 − x
2
x + ax + b< 0
has the solution 4 ≤ x< 5, find the conditions a and b should satisfy.
⎧ 2 ⎧
⎨ x − 2x − 8 ≥ 0 ⎨ x ≤−2 or x ≥ 4
Solution: √ 2 ⇒ x ≤−2
x − 2x − 8 < 8 − x ⇒ 8 − x> 0 ⇒ x< 8
⎩ 2 2 ⎩ 36
x − 2x − 8 < (8 − x) x<
7
36
or 4 ≤ x ≤ 7 .
Download free eBooks at bookboon.com
107