Page 103 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.47 Consider the inequality x +2 >m(x − 1), (1) if the inequality holds for any real
2
number x , find the range of m ; (2) if for any m ∈ [−2, 2] the inequality holds, find the

range of x .

Solution: (1) x +2 >m(x − 1) ⇔ m(x − 1) − (x + 2) < 0 ⇔ mx − x − m − 2 < 0. This
2
2
2

m< 0
m< 0
m< 0

inequality holds for any real number x , then m< 0 ⇔ m< 0 ⇔ m< √0 √ ⇔
2
Δ = 1 +4m(m + 2) < 0 ⇔ 4m +8m +1 < 0 ⇔ −1 − 3 √ <m< −1+ 3 √ 3 ⇔
2
3
√ Δ = 1 +4m(m + 2) < 0 4m +8m +1 < 0 −1 2 − 2 <m< −1+ 2 2
√
m< 0 m< 0 m< 0 3 √ 3 √
3
−1 −
<m < −1+
⇔ ⇔ √ √ ⇔ −1 2 − <m < −1+ 2 3 .
2
Δ = 1 +4m(m + 2) < 0 4m +8m +1 < 0 −1 − 3 <m< −1+ 3 2 2
√ √ 2 2
3 3
2
−1 − <m < −1+ (2) For any m ∈ [−2, 2] the inequality holds. Let f(m) =(x − 1)m − (x + 2) which is a
2 2
linear function of m , and f(m) < 0 should hold for any m ∈ [−2, 2], equivalently we
2 2 2 2 1 1
−2(x − 1) − (x + 2) < 0
2x + x> 0
should have −2(x − 1) − (x + 2) < 0 ⇔ 2x 2 2x + x> 0 ⇔ x> x> 0 or x< − 2 √ 1+ 33 ⇔ ⇔
x> 0 or x< −

1
2
−2(x − 1) − (x + 2) < 0
⇔+ x> 0
2 √
√
⇔ 0 or x< −
√
2
2
1− 33
2
2x − x − 4 < 0
2
2(x − 1) − (x + 2) < 0
1− 33
1+ 33
2 √
√
2(x − 1) − (x + 2) < 0 ⇔
2x − x − 4 < 0 ⇔
2
2
< x< 33
1+
2(x − 1) √ − √ (x + 2) < 0 2x − x − 4 < 0 1− 33 4 < x< < x< ⇔ 4
4
4
0 < x< 1+ 33 √ 4 4
1+ 33
√
0 < x< 33
1
0 < x< 1+ 4 or 4 1− 33 < x< − .
4 4 2
3.48 x,y,z are positive numbers, and xyz(x + y + z) =1. Find the minimum value
of (x + y)(x + z).
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