Page 121 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

5
1
1
2
1
1
5
x + 3 3 2 2 2 5 5 3 3 1 2 2 3 3 1 2 2 3 3 x · 5 3 3 1 2 3 3 1 2 2 3 3
√ x = x .
√ x ≥ 3
√ x +
√ x = x +
5
√ x · 2
√ x = x +
√ x = x
√ x ≥ 3
√ x +
x +
x ·
√ x ·
3 3
3 3
3 3
3 3
3 3
2
2
5
3
2
3
2
5
Similarly, we can obtain y + √ y ≥ y ,z + √ z ≥ z .
3 3 3 3
2
2
5
5
3
3
2
2
5
3
Add these three inequalities up to obtain x + y + z + √ (x + y + z ) ≥ x + y + z .
3 3
2
3
3
5
5
3
5
2
2
2
Since x + y + z =1, then x + y + z + √ ≥ x + y + z ,
3 3
3 2 3 2 3 2 √ (i).
2
then x (1 − x )+ y (1 − y )+ z (1 − z ) ≤
3 3
y
2

3
3
z
2
2
3
3
2
y
x
3
[x (1 − x )+ y (1 − y )+ z (1 − z )]( 2 x 2 + 1−y + 2 + 1−z + 2 ) ≥ ( x (1 − x ) 2 1−x 2 + 2 +
3
z
3
x
2
2
x
3
2 ) ≥ (
[x (1 − x )+ y (1 − y )+ z (1 − z
1−x )](
x (1 − x )
1−x 2 1−y 2 1−z 1−x

y
2
2
2
3
y (1 − y ) 2 1−y 2 + 2 + z (1 − z ) 2 1−z 2 = x + y + z =1
3
z
2
2
y
z
2
2 = x + y + z =1 . Thus
2
2
3
3
z (1 − z )
y (1 − y )
1−y
1−z
x x 2 + + y y 2 + + z z 1 1
2 2
3
3 x (1−x )+y (1−y )+z (1−z
2
2 2
3 3
2 2
3 3
1−x
1−y
1−z
1−x 2 1−y 2 1−z 2 ≥ ≥ x (1−x )+y (1−y )+z (1−z ) ) (ii). From (i) and (ii), we obtain
√
x y z 3 3
1−x 2 + 1−y 2 + 1−z 2 ≥ 2 .
√
1 x y z 3 3
When x = y = z = √ , 2 + 2 + 2 reaches the minimum value .
3 1−x 1−y 1−z 2
3.93 Positive numbers a 1 ,a 2 , ··· ,a n and b 1 ,b 2 , ··· ,b n
satisfy a 1 + a 2 + ··· + a n ≤ 1,b 1 + b 2 + ··· + b n ≤ n ,
n
show ( 1 + 1 )( 1 + 1 ) ···( 1 + 1 ) ≥ (n + 1) .
a 1 b 1 a 2 b 2 a n b n
) ≤
Proof: The given conditions together with Mean Inequality result in a 1 a 2 ··· a n ≤ ( a 1 +a 2 +···+a n n 1
n n n

n

(i), and b 1 b 2 ··· b n ≤ ( b 1 +b 2+···+b n )=1 (ii). In addition, 1 + 1 = 1 + ··· + 1 + 1 ≥ (n + 1) n+1 1 1
n
a i b i na i na i b i na i b i

n

1 1 1 1 1 1 1 n terms
+ = + ··· + + ≥ (n + 1) n+1 (i =1, 2, ··· ,n ) (iii).
a i b i na i na i b i na i b i

n terms
From (i),(ii),(iii), we can obtain

n
1 1 1 1 1 1 1 1 1
+ + ··· + ≥ (n + 1) n n+1 · ·
n n
a 1 b 1 a 2 b 2 a n b n (n ) a 1 a 2 ··· a n b 1 b 2 ··· b n

1
n n
≥ (n + 1) n n+1 · (n ) · 1
n n
(n )
n
=(n + 1) .
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