Page 33 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

x y z xyz
3
Solution: Let = = = t ⇒ = t , and x + y y + z z + x (x + y)(y + z)(z + x) (a + b)(b + c)(c + a)
a b c abc = = = t ⇒ = t ⇒ =
3
a + b b + c c + a (a + b)(b + c)(c + a) (x + y)(y + z)(z + x)
x + y y + z z + x (x + y)(y + z)(z + x) (a + b)(b + c)(c + a)
3
= = = t ⇒ = t ⇒ = 1
a + b b + c c + a (a + b)(b + c)(c + a) (x + y)(y + z)(z + x) 3 .
1 t
t 3
Hence,
xyz(a + b)(b + c)(c + a) 1
= t 3 =1.
abc(x + y)(y + z)(z + x) t 3
1 1 1
1.88 Given 2007x = 2009y = 2011z ,x > 0,y > 0,z > 0, and + + =1,
2
2
2
x y z
√ √ √ √
show 2007x + 2009y + 2011z = 2007 + 2009 + 2011.
Proof: Let 2007x = 2009y = 2011z = k (k> 0), then
2
2
2
1 1 1
2007x = k/x, 2009y = k/y, 2011z = k/z . Since + + =1, then
x y z
√
2007x + 2009y + 2011z = k/x + k/y + k/z = k(1/x +1/y +1/z) = k.
On the other hand,
√ √ √ √
2 2 2 √ √ √ √
2007 = k/x , 2009 = k/y , 2011 = k/z ⇒ 2007 + 2009 + 2011 = k/x +
√ 2
2
2
√
√ 2007 = k/x , 2009 = k/y , 2011 = k/z ⇒ 2007 + 2009 + 2011 = k/x +
√
√
k/y + k/z = k √
k/y + k/z = k . Hence the aimed equality holds.
1.89 If x,y,z are nonzero real numbers, and x + y + z = xyz, x = yz , show x ≥ 3.
2
2
Proof: x + y + z = xyz ⇔ y + z = xyz − x = x − x since yz = x . Then we can treat
2
3
y, z as two roots of the quadratic equation u − (x − x)u + x =0. . Since x,y,z are real
2
2
3
numbers, then
3 2 2 6 4 2 2 4 2
Δ=(x − x) − 4x ≥ 0 ⇒ x − 2x − 3x ≥ 0 ⇒ x (x − 2x − 3) ≥ 0 ⇒
2
2
2
6
4
2
2
3
4
Δ=(x − x) − 4x ≥ 0 ⇒ x − 2x − 3x ≥ 0 ⇒ x (x − 2x − 3) ≥ 0 ⇒
2
2
2
x (x + 1)(x − 3) ≥ 0
x (x + 1)(x − 3) ≥ 0. Since x =0, then x > 0,x +1 > 0, thus x − 3 ≥ 3 0, i.e. x ≥ 3.
2
2
2
2
2
2
2
1.90 Given a, b, c are nonzero real numbers, and a + b + c =1 ,
2
2
2
1
1
1
1
1
a + b + c )+ b( + )+ c( + ) + 3 = 0, find all possible values of a + b + c .
1
( +
b c c a a b
Solution:
1
1
1
1
1
1
( +
a + b + c )+ b( + )+ c( + ) +3 = 0 ⇒ ac+ab + ab+bc + bc+ca +3 = 0 ⇒ (a + b +c)
b c c a a b bc ac ab
1
1
1
1
1
1
( + )+ b( + )+ c( + ) +3
c)(ab + bc + ca)=0 ⇒ a + b + c =0 = 0 ⇒ ac+ab + ab+bc + bc+ca +3 = 0 ⇒ (a + b +
ab
c
b
bc
a
b
a
c
ac
c)(ab + bc + ca)=0 ⇒ a + b + c =0 or ab + bc + ca =0. For the case ab + bc + ca =0,
since a + b + c =(a + b + c) − 2(ab + bc + ca) =1, then
2
2
2
2
(a + b + c) =1 ⇒ a + b + c = ±1 . Hence, a + b + c can be -1, 0, or 1.
2
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