Page 114 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

1 + 1 + 1 ≥ 3 3 1 , (a+1)+(b+1)+(c+1) ≥ 3 3 (a + 1)(b + 1)(c + 1)
a+1 b+1 c+1 (a+1)(b+1)(c+1) 1 1 1 1 3
Therefore, ( + + )[(a+1)+(b+1)+(c+1)] ≥ 3 3 ·3 (a + 1)(b + 1)(c + 1) =
a+1 b+1 c+1 (a+1)(b+1)(c+1)
( 1 + 1 + 1 )[(a+1)+(b+1)+(c+1)] ≥ 3 3 1 ·3 3 (a + 1)(b + 1)(c + 1) = 9 . Since
a+1 b+1 c+1 (a+1)(b+1)(c+1)
9
1 + 1 + 1 9 9 = 3
0 <a + b + c ≤ 3, then a+1 b+1 c+1 ≥ (a+1)+(b+1)+(c+1) ≥ 3+3 2 .
3.79 Given a, b, c, m, n, p > 0, and a + m = b + n = c + p = R , show an + bp + cm < R 2

an + bp + cm < R .
2
Proof: Construct an equilateral triangle ABC with side length R . Choose points

D,E,F on sides AB, BC, CA respectively such that

AD = a, DB = m, BE = c, EC = p, CF = b, FA = n . In this way, three side lengths are

a + m, c + p, b + n , and a + m = c + p = b + n = R . Connect D with E , connect E

with F , and connect F with D . Let S ADF = S 1 ,S BDE = S 2 ,S CEF = S 3 ,S ABC = S

√
1
1
1
0
0
0
S ADF = S 1 ,S BDE = S 2 ,S CEF = S 3 ,S ABC = S . Then S 1 + S 2 + S 3 = an sin 60 + cm sin 60 + bp sin 60 = 3 (an + cm + bp)
2 2 2 4
√ √
1
1
1
1
0
0
2
0
0
2
S 1 + S 2 + S 3 = an sin 60 + cm sin 60 + bp sin 60 = 3 (an + cm + bp). S = R sin 60 = 3 R . S = S 1 + S 2 + S 3 + S DEF >S 1 + S 2 + S 3 , thus
2 2 2 4 2 4
√ √
3 (an + cm + bp) < 3 R , that is, an + bp + cm < R .
2
2
4 4
2
3.80 Let x 1 ,x 2 , ··· ,x n are positive numbers, show x 2 1 + x 2 2 + ··· + x 2 n−1 + x n ≥ x 1 + x 2 + ··· + x n
x 2 x 3 x n x 1
x 2 x 2 x 2 2
1 + 2 + ··· + n−1 + x n ≥ x 1 + x 2 + ··· + x n .
x 2 x 3 x n x 1
2 2 2 x 2 1
Proof 1: Since x 1 ,x 2 , ··· ,x n > 0 , we can do the following: (x 1 − x 2 ) ≥ 0 ⇒ x + x ≥ 2x 1 x 2 ⇒ x 2 + x 2 ≥ 2x 1

2
1
2 2 2 x 2 1 x 2 2 x 2 n−1 x n 2
(x 1 − x 2 ) ≥ 0 ⇒ x + x ≥ 2x 1 x 2 ⇒ x 2 + x 2 ≥ 2x 1. Similarly, we can obtain x 3 + x 3 ≥ 2x 2 , ··· , x n + x n ≥ 2x n−1 , x 1 + x 1 ≥ 2x n
2
1
2
2
x
2
x
x
. Add them up to obtain ( x 2 1 + x 2 2 + ··· + x 2 n−1 ( + 1 2 x n 2 ) +(x 1 + x 2 + ··· + x n ) ≥ 2(x 1 + x 2 + ··· + x n ) ⇒
+
+ ··· +
+
n−1
2
x n
) +(x 1 + x 2 + ··· + x n ) ≥ 2(x 1 + x 2 + ··· + x n ) ⇒
x 2 2 x 3 2 x n x 2 x 1 x 3 x n x 1
2
x
x
x
+ ··· +
2
n−1
1
≥ x 1 + x 2 + ··· + x n ≥ x 1 + x 2 + ··· + x n.
( x 2 1 + x 2 2 + ··· + x 2 n−1 + x n 2 ) +(x 1 + x 2 + ··· + x n ) ≥ 2(x 1 + + x 2 + ··· + x n ) ⇒ + x x 1 2 2 n + x 2 2 + ··· + x 2 n−1 + x 2 n
x 1
x 2 x 3 x n x 1 x 2 x 3 x n x 2 x 3 x n x 1
x 2 1 + x 2 2 + ··· + x 2 n−1 + x 2 n
Proof 2: Mean Inequality implies that 1 + x 2 ≥ 2 1 · x 2 =2x 1 . Similarly, we have
x 2 x 3 x n x 1 ≥ x 1 + x 2 + ··· + x n x 2 x 2
x 2 x 2
2
x 2 2 + x 3 ≥ 2x 2 , ··· , x 2 n−1 + x n ≥ 2x n−1 , x n + x 1 ≥ 2x n . Add them up to obtain the result.
x 3 x n x 1
2
2
2
2
2
Proof 3: x n + x 2 1 + x 2 1 + x 2 2 + ···+ x 2 n−1 + x n = 2 x n +x 2 1 + x +x 2 2 + ···+ x 2 n−1 2 +x n ≥ 2( x 1x n + x 1 x 2 +
1
2
2
2
2
2
2
2
2
2
2
x
x
+
+
=
+
+
x 1
1
x 2
x n + x n−1 x n + x 2 + ···+ x n−1 x n x n x n x n +x 1 x 1 + x +x 2 x 2 + ···+ x n−1 +x x n n ≥ 2( x 1x n x 1 x 1 x 2 x 2
x 1
1
x 2
1
x n
x n
2 2 2 2 x 2 2 2 2 2 2 x 2 2 ··· + x 1 x 2 ) = 2(x 1 + x 2 + ··· + x n ) x 1 x 2 x n x 1 x 2
x 1
x 2
+ + + + ···+ + = + + ···+ ≥ 2( + + ··· + x n ) = 2(x 1 + x 2 + ··· + x n ) which implies the result.
1
x n x 1 x 1 x 2 n−1 x n x n +x 1 x +x 2 n−1 +x n x 1x n x 1 x 2 x n−1 x n
x 1 x 1 x 2 x 2 x n x n x 1 x 2 x n x 1 x 2 x n
··· + x n−1 x n ) = 2(x 1 + x 2 + ··· + x n )
x n
Proof 4: x 2 1 + x 2 2 + ··· + x 2 n−1 + x n 2 = [x 2−(x 2 −x 1)] 2 + [x 3−(x 3 −x 2)] 2 + ··· + [x n−(x n−x n−1)] 2 +.
x 2 x 3 x n x 1 x 2 x 3 x n
[x 1−(x 1 −x n)] 2
=(x 2 + x 3 + ··· + x n + x 1 ) − 2(x 2 − x 1 + x 3 − x 2 + ··· + x n − x n−1 + x 1 −
x 1 2 2 2 2 2
x n )+ (x 2 −x 1) + (x 3 −x 2) + ··· + (x n−x n−1) + (x 1 −x n) =(x 1 + x 2 + ··· + x n )+ (x 2 −x 1) +
x 2 x 3 x n x 1 x 2
(x 3 −x 2) 2 (x 1 −x n) 2
+ ··· + ≥ x 1 + x 2 + ··· + x n
x 3 x 1
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