Page 46 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1
+yz + zx − 2xyz = [(1 − 2x)(1 − 2y)(1 − 2z) + 1]
4 .
Since x + y + z =1 , then at most one of 1 − 2x, 1 − 2y, 1 − 2z is less than zero, thus

3 3
1 − 2x +1 − 2y +1 − 2z 3 3 − 2(x + y + z) 3

(1 − 2x)(1 − 2y)(1 − 2z) ≤ 1 − 2x +1 − 2y +1 − 2z = 3 − 2(x + y + z) =
(1 − 2x)(1 − 2y)(1 − 2z) ≤ 3 = 3 =
3 3 3
3 − 2 3 1

1
3 − 2 = 1 . Hence, xy + yz + zx − 2xyz≤ ( 1 + 1) = 7 . Therefore, the maximum value
+yz + zx − 2xyz
3 27 = 4 27 27
3 27
of xy + yz + zx − 2xyz is 7/27.
1.120 Let a, b, c ∈ R,a + b + c =0, show
2
5
2
3
3
5
a + b + c 5 a + b + c 2 a + b + c 3
= · .
5 2 3

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