Page 94 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

x+y
3.16 If −1 < x< 1, −1 <y < 1, show | 1+xy | < 1 .
x+y x+y 2 2 2 2 2 2 2
x+y
x+y
2
2
2 2
2
2
Proof: | | < 1 ⇔ ( 2 ) < 1 ⇔ (x + y) < (1 + xy) ⇔ x + y < 1+ x y ⇔
1+xy < 1 ⇔ (
| 1+xy | 1+xy )
1+xy < 1 ⇔ (x + y) < (1 + xy) ⇔ x + y < 1+ x y ⇔
2
2
(x − 1)(1 − y ) < 0
(x − 1)(1 − y ) < 0, which is obviously valid since −1 < x< 1, −1 <y < 1.
2
2
x
3.17 Given f(x)=lg 1+2 +a·4 x (a ∈ R ), (1) f(x) is well defined when x ≤ 1, find
3
the range of a , (2) if 0 <a ≤ 1, show 2f(x) <f(2x) when x =1.
x
Solution: (1) Since 1+2 + a · 4 > 0, a> −[( ) +( ) ]. Since ( ) , ( ) are decreasing
x
1 x
1 x
1 x
1 x
4 2 4 2
1 x
1 x
functions on the interval (−∞, 1] , then −[( ) +( ) ] reaches the maximum value
4 2
1
1
−( + ) = − at x =1, thus a> − .
3
3
4 2 4 4
x x x x
x
x 2
x
2
x
x
2
2
(2) Use the inequality a+b+c < a +b +c 2 to obtain (1+2 +a·4 ) < 3(1+4 +a ·16 ) < 3(1+4 +a·16 ) ⇒ 1+4 +a·16 > ( 1+2 +a·4 ) 2
x
3 3 3 3
x
x
x x 2 x 2 x x x 1+4 +a·16 x 1+2 +a·4 x 2
(1+2 +a·4 ) < 3(1+4 +a ·16 ) < 3(1+4 +a·16 ) ⇒ > ( ) , that is f(2x) > 2f(x).
3 3
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