Page 101 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

Proof: If one of x,y,z is zero, without loss of generality, assume x =0, then |y + z| =0,
thus y + z =0, which implies x + y + z =0. If x,y,z are all nonzero, then there are four
possibilities:

1) If x,y,z are all positive, then y + z ≤ x, z + x ≤ y, x + y ≤ z , impossible.
2) If x,y,z have two positive one negative, without loss of generality, assume
x> 0,y > 0,z < 0. |y + z|≤ x ⇒−x ≤ y + z ≤ x ⇒ x + y + z ≥ 0. On the
other hand |x + y|≤ |z|⇒ x + y ≤−z ⇒ x + y + z ≤ 0. As a conclusion,
x + y + z =0.
3) If x,y,z have one positive two negative, without loss of generality, assume
x> 0,y < 0,z < 0 . |y + z|≤ x ⇒ y + z ≥−x ⇒ x + y + z ≥ 0. On the other
hand, |x + y|≤ |z|⇒ x + y ≤−z ⇒ x + y + z ≤ 0. As a conclusion,
x + y + z =0.
4) If x,y,z are all negative, then x ≤ y + z ≤−x, y ≤ z + x ≤−y, z ≤ x + y ≤−z ,
then x + y + z ≤ 2(x + y + z), thus x + y + z ≥ 0, a contradiction to the assumed
negativity condition.

3.41 If x> y > 0, show x − y + 2xy − y >x .
2
2
2
2 2 2 2 2 2 2 2
2
2
2
2
2
2
2
2
Proof 1: x>y > 0 ⇒ xy > y , 2xy − y >y ⇒ x − y >x − 2xy + y =(x − y) ⇒
x>y > 0 ⇒ xy > y , 2xy − y >y ⇒ x − y >x − 2xy + y =(x − y) ⇒

2
2
x − y >x − y x − y + 2xy − y >x − y + y = x .
2
2
2
2
2
2
x − y >x − y , and 2xy − y >y , thus

2 2 2 x − y >x − y
2
2
Proof 2: x> y > 0 ⇒ y> −y ⇒ x + y >x − y ⇒ x − y > (x − y) ⇒

2 2 2 2xy − y >y (ii).
2
(i). 2xy > 2y ⇒ 2xy − y >y ⇒

2
2
2
(i)+(ii)⇒ x − y + 2xy − y >x .

2
2
2
2
2
2

2
2
2

2
Proof 3: x − y + 2xy − y >x ⇔ x −y +2 2 (x − y )(2xy − y ) +2xy −y >x ⇔
2
2
2
2
2
2
2
2
2
x − y +
2xy − y >x ⇔ x −y +2
(x − y )(2xy − y ) +2xy −y >x ⇔
(x − y )(2xy − y ) >y − xy . The left hand side is greater than zero, while he right
2 2 2 2
(x − y )(2xy − y ) >y − xy
2
2
2
2

hand side y − xy = y(y − x) < 0, thus (x − y )(2xy − y ) >y − xy always holds.
2
2
2
2
2
1
3.42 Given x> 0,y > 0, + 9 y =1, show x + y ≥ 12.
x

Proof: Since x> 0,y > 0, we have 1 x + 9 y ≥ 2 1 x · 9 y = √ 6 xy . Since 1 x + y 9 =1 , we have
6 √ √
√ ≤ 1 , which is equivalent to xy ≥ 6. x + y ≥ 2 xy ≥ 12.
xy
3.43 a, b, c are real numbers and a + b + c =1, show a + b + c ≥ .
2
2
2
1
3
1
1
1
2
2
2
2
2
2
2
2
2
2
2
Proof 1: a + b + c =1 ⇒ c =1 − a − b , then a +b +c − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a +
1
2
2
2
2
2
2
2
1
2

2
2
1
2
a +b +c 3 − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a +
3
3
3
2 1 3 2 b−1 2 3 b−1 2 2 1 b−1 2 (3b−1) 2 2 ] ≥ 0
) + (3b−1)
1
1
1
2
2
2
1
2
2
2
2
2
2
2
1
2
2
2
) +b −b+ ] = 2[(a+ b−1 2
) −( b−1 2
b +ab−a−b+ ) = 2[a +(b−1)a+( b−1 2
2
2
1
2
2
2
2
2
2
2
2
2
2
1
a +b +c − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a + b +ab−a−b+ 3 ) = 2[a +(b−1)a+( 2 ) −( 2 ) +b −b+ ] = 2[(a+ 2 ) + 12 12 ] ≥ 0
1

3
a +b +c
3 − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a +
3
3
2
3
2
3
2
3
3
3
2
b−1 2
2
1
2
) +
(3b−1)
) +b −b+ ] = 2[(a+ b−1 2
2
b +ab−a−b+ ) = 2[a +(b−1)a+( b−1 2 b−1 2 2 2 1 3 1 3 b−1 2 (3b−1) 2 ] ≥ 0 .
2
1
) −( b−1 2
2 ) +
2 ) +b −b+ ] = 2[(a+
] ≥ 0
3 ) = 2[a +(b−1)a+(
2 ) −(
b +ab−a−b+
12
12
2
2
3
2
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