Page 116 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.81 If x, y are real numbers, and y ≥ 0,y(y + 1) ≤ (x + 1) , show y(y − 1) ≤ x .
2
2
2
2
2
2
1 2
2
Proof: If 0 ≤ y ≤ 1, obviously y(y − 1) ≤ 0 ≤ x . If y> 1, then y(y + 1) ≤ (x + 1) ⇒ y + y + 1 ≤ (x + 1) + 1 ⇒ (y + ) ≤ (x + 1) + 1 ⇒ 1 <
2
2
2
4
2
1 2
2
4
y(y + 1) ≤ (x + 1) ⇒ y + y + 1 ≤ (x + 1) + 1 ⇒ (y + ) ≤ (x + 1) + 1 ⇒ 1 < 4
2 1 1 4 4 2 4

−
1
2
2
2
1 2
2
2
(x + 1) +
y(y + 1) ≤ (x + 1) ⇒ y + y + 1 ≤ (x + 1) + 1 ⇒ (y + ) ≤ (x + 1) + 1 ⇒ 1 < y ≤ y ≤ (x + 1) + 1 2 4 . The inequality
2
−
4
4 4 2 4
2
y ≤ (x + 1) + 1 − 1 to prove 2 2 2 2 2 2 1 1 1 2 2 2 1 1 1 ⇔ (y − ) ≤ x + 1 1 2 1 1 1 1 1 1
1 2 2
2
1 2
1
22 1
2 2
1
1
1
2
y(y − 1) ≤ x ⇔ y − y + ≤ x + ⇔ (y − ) ≤ x + 2
2
1 2
4 2 y(y − 1) ≤ x ⇔ y − y + 4 + 1 4 x + 2 4 1 4 4⇔ (y − ) ≤ x + 4 4 ⇔ y ≤ x + x + + 2 ⇔ ⇔
x
4+
y(y − 1) ≤ x ⇔ y − y
4 2 +
4 ≤ x +
⇔ y ≤ y ≤ x + 2
⇔
+ + ⇔
y(y − 1) ≤ x ⇔ y − y + ≤
2 2
2 ⇔
≤ x + ⇔ (y − ) ≤ x + ⇔ y ≤
2
4
4
4
2
4
2
1
2
1
1
2

1− y +
2 ) ≤ x +
y(y − 1) ≤ x 2 1 ⇔ y 1 4 1 ≤ x + 1 2 2 4 x + 2 + 1 2
⇔ (y −
1
(x + 1) + − x + + 1 1 ⇔ 4 (x + 1) + 1 ≤ 2 ⇔ y ≤ 4 2 1 ⇔
1
1
≤ 1
1
2 1
2 2
2 1
2 1 1
2 1
2 4
≤ 2 14
2 1 1
x + +1 ⇔ (x + 1) + 2 1 2
2
2
1
1
1
x + 1
1
4 +1
x
4 +
(x + 1)
x2
(x + 1)
2
4 2 +
2
x
4 +
(x + 1) + ≤
(x + 1) + − 4 ≤ x + + 1 2 ⇔ (x + 1) + 2 4 ≤ x + +1 ⇔ (x + 1) + 2 4 1 ≤
(x + 1) + − ≤
4 + +1 ⇔ (x + 1) + ≤
2 ⇔
2 ≤
4 4 2 − 2 2 4 + 4 4 + ⇔ 4 4 + 4 ⇔ (x + 1) + 4 ≤
4
4 ≤
2
4
1
1
1
1
1
2 2
1 2
1 2
1 2
2 2
x + +1
x + +

(x + 1)
(x + 1) + 1 x + +1 ⇔ x +2x +1 ≤ x +2 + x + +1 ⇔ x ≤ ⇔ (x + 1) + 1 ≤

1
2
1
⇔
2 ≤
x +
2 ≤
2
1
1
1
1
2 1 2
2 2 1 x + +2 − 1 1 4 2 2 4 2 1 1 1 4 2 2 4 2 2 1 4 +1 ⇔ x ≤ x + 4 + 1 1 4
2 1
2 4
4 1 4
2
1
x + 1
1
2
2
x
2
4 +
4 +1 ⇔ x +2x +1 ≤ x +2
x 2 4 +2 4 x + +1 ⇔ x +2x +1 ≤ x +2 x + +1 ⇔ x ≤ x
x
x2
2
4
x + +2 x + +1 ⇔ x +2x +1 ≤ x +2
x + +2 4 + 4 4 4 4 4 + 4 +1 ⇔ x ≤ 4 + 4 4
4
4
4
4
1
1
2
1
2
1
2
1
2
2
2
x + +2 x + +1 ⇔ x +2x +1 ≤ x +2 x + +1 ⇔ x ≤ x + 1 which is
4 4 4 4 4 4
obviously valid.
3.82 If real numbers x,y,z satisfy x + y + z =2, show x + y + z ≤ xyz +2.
2
2
2
Proof: If one (or more) of x,y,z is not positive, without loss of generality let z ≤ 0. Since
1 2 2 1 2 2 2
2
2
2
2
2
x+y ≤ 2(x + y ) ≤ 2(x + y + z ) =2, xy ≤ (x +y ) ≤ (x +y +z )= 1, then
2 2
2+ xyz − (x + y + z) = [2 − (x + y) − z(xy − 1)] ≥ 0, that is, x + y + z ≤ xyz +2.
If x,y,z are all positive, let 0 <x ≤ y ≤ z .
When z ≤ 1, 2+ xyz − (x + y + z)=1 − x − y + xy +1 − xy − z + xyz = (1 − x) − y(1 − x)+

2+ xyz − (x + y + z)=1 − x − y + xy +1 − xy − z + xyz = (1 − x) − y(1 − x)+
(1 − xy) − z(1 − xy) = (1 − x)(1 − y) + (1 − xy)(1 − z) ≥ 0

2+ xyz − (x + y + z)=1 − x − y + xy +1 − xy − z + xyz = (1 − x) − y(1 − x)+ (1 − xy) − z(1 − xy) = (1 − x)(1 − y) + (1 − xy)(1 − z) ≥ 0 , that is,x + y + z ≤ xyz +2

(1 − xy) − z(1 − xy) = (1 − x)(1 − y) + (1 − xy)(1 − z) ≥ 0
x + y + z ≤ xyz +2.
√
2
2

When z> 1, x + y + z ≤ 2[z +(x + y) ]= 2(2 + 2xy)= 2 1+ xy ≤ 2+ xy < 2+ xyz
√
2
2
x + y + z ≤ 2[z +(x + y) ]= 2(2 + 2xy)= 2 1+ xy ≤ 2+ xy < 2+ xyz .
As a conclusion, x + y + z ≤ xyz +2 holds.
3.83 Given the function f(x) = ax + bx + c (a> 0), and the two roots of the
2
1
equation f(x) − x =0 satisfy 0 <x 1 <x 2 < . (1) When x ∈ (0,x 1 ), show x< f(x) <x 1 ;
a
(2) Assume the curve of the function f(x) is symmetric about the straight line x = x 0, show
x 0 < x 1 .
2
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