Page 110 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

2
x − x − 2 > 0
3.67 If the system of inequalities 2x +(5+2k)x +5k< 0 has only one integer
2
solution −2, find the range of k .

Solution: The solution of x − x − 2 > 0 is x< −1 or x> 2. The second inequality is
2
equivalent to (2x + 5)(x + k) < 0. When −k< − , i.e. k> , the second inequality has
5
5
2 2
5
5
the solution −k <x < − , in which −2 is not included. When −k> − , i.e. k< , the
5
2 2 2
second inequality has the solution − <x< −k , then the solution of the inequality system
5
2
x> 2
is x< −1 or − <x < −k . To have only one integer solution −2, we should
5
5
− <x < −k 2
2
5
have −k ≤ 3 and −k> −2, that is, −3 ≤ k< 2. When −k = − , i.e. k = , the second
5
2 2
inequality has no solution. As a conclusion, k ∈ [−3, 2).
3 .
a b c
3.68 Let a, b, c are positive numbers, show a b c ≥ (abc) a+b+c
3 , we only need
a b c
Proof: Without loss of generality, let a ≥ b ≥ c> 0. To show a b c ≥ (abc) a+b+c
b−c ≥ 1, we only need to show
3a 3b 3c
to show a b c ≥ (abc) a+b+c , we only need to show a a−b b b−a c c−a
a c−a b c−a c
a a−b b b−c c a−c a b a ≥ 1, thus the
a−c ≥ 1. Since a − b ≥ 0,b − c ≥ 0,a − c ≥ 0, we have ≥ 1, ≥ 1,
b a−b c b−c a b c c
last inequality holds.
2
2
2
2
2
2
3.69 If a,b,c,x,y,z are all real numbers, and a + b + c = 25,x + y + z = 36, ax + by + cz = 30

a+b+c
a + b + c = 25,x + y + z = 36, ax + by + cz = 30, find the value of x+y+z .
2
2
2
2
2
2
Solution: Cauchy’s Inequality implies
25 × 36 = (a + b + c )(x + y + z ) ≥ (ax + by + cz) = 30× 36 = (a + b + c )(x + y + z ) ≥ (ax + by + cz) = 30 . The equal sign is
25 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
obtained since 25 × 36 = 30 . Thus there exist λ, μ (not both zero) such that
2
λa = μx, λb = μy, λc = μz . Therefore
2 2
2 2
2 2
λ (a + b + c )= μ (x + y + z ) ⇒ 25λ = 36μ ⇒ 5λ = ±6μλ (a + b + c )= μ (x + y + z ) ⇒ 25λ = 36μ ⇒ 5λ = ±6μ . However,
2 2
2 2
2 2
2 2
2 2
2 2
2 2

μ 5 a+b+c μ 5
ax + by + cz = 30, thus 5λ =6μ ⇒ λ = 6 ⇒ x+y+z = λ = .
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