Page 115 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

√ √ √ √
Proof 5: Since x 1 ,x 2 , ··· ,x n > 0, let a 1 = x 2 ,a 2 = x 3 , ··· ,a n−1 = x n ,a n = √ ,b 2 = x 2 √ ,b 3 =
x 1
√
√
x 2
√
√
x 2 =
x 3 =
a 1 = x 2 ,a 2 = x 3 , ··· ,a n−1 = x n ,a n = x 1 ,b 1 = √ ,b 2 √ ,b 3
x 1 ,b 1 = x 1
x 3 x n−1 x n x 2 x 3
√ √ √ √ √ , ··· ,b n−1 = √ ,b n = √
x n−1
x 3
a 1 = x 2 ,a 2 = x 3 , ··· ,a n−1 = x n ,a n = x 1 ,b 1 = √ ,b 2 = √ ,b 3 = √ , ··· x 4 ,b n−1 = √ ,b n x n = √ x n . Cauchy Inequality implies that
x 2
x 1
x 1
x 2 x 3 x 4 x n x 1
x 3 x n−1 x n
√ , ··· ,b n−1 = √ ,b n = √ √ 2 √ 2 √ 2 √ 2 2 2 x n−1 2 2
x n
x 4 x n x 1 2 2 2 2 2 2 2 √ √ √ √ x 1 x 2 ) +( √ ) ] ≥
(a + a + ··· + a )(b + b + ··· + b ) ≥ (a 1 b 1 + a 2 b 2 + ··· + a n b n ) , then [( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √ ) +···+( √ 2
2
2
2
2
2
x n−1 2
2
√
√
x n
x 2
√
x 1
√
1 2 n 1 2 n [( x 2 ) +( x 3 ) 2 +···+( x n ) +( x 1 ) ]·[( √ x 1 x 2 ) +( √ ) +···+( √ x n ) +( √ ) ] ≥
x n−1 2
2
x 3 2
2
2
x 1 2
2
x 2
x n
2 x 1) ] ≥
√ [( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √ ) +···+( √ x n) +( √ x x 1 x 2
√
x 3
√
x 2
√
2
x n−1
√ √ 2 √ √ 2 2 2 x n−1 2 2 √ √ x 1 + √ x 3 x 2 √ x 2 + ··· + √ x n x n−1 + √ x 1 x n x n x 2 x 3 x n 2 2 1 + x 2 2 2 +
√ ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x
√
2
√
x n
[ x 2 x 1
x 1

x 2

x
2 √ √
2 √ √
√ √
√ √
2
√
√
√
√
x
x
[( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √ ) +···+( √ ) +( √ ) ] ≥ [ x 2 x 1 x 2 + √ x 3 x 2 x 3 + ··· + √ x n n−1 x n + √ x 1 x n x 1 ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x 1 1 x 2 + x 2 2 x 3 +
x
2 2
2 2
2 2
2 2
2 2
x n−1 2 2
2 2
2 2
x n−1
x 3 ) +···+(
x 1
x n n
x 2 2
[( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ 1 x 2 ) +( √ ) +···+( √ √ x n ) +( √ ) ] ≥ [ x 2 √ x 2+ x 3 √ x 3+ ··· + x n √ x n + x 1 √ x 1] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x 2+ x 3+
2
x 1 ) ] ≥
[( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √
2
x
) +( √
n
≥ (x 1 + x 2 + ··· + x n−1 + x n )
x 3 3
x 2 2
x n n
√ √ √ √ x x x x x x x 2 ··· + x x 2 2 n−1 + x 2 x x 3 x n x 1 2 x 2 x 3
2 1 1
2
+
x n
√
√
√ √ √ x 1 + √ √ x 2 + ··· + √ x n−1 √ √ √ ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x x 2 2 1 + x x 2 2 2 x 2 n−1 2 n x 1 ≥ (x 1 + x 2 + ··· + x n−1 + x n ) 2 2
x n +x
x
2 2

√
x
+ ··· + n−1
x n n−1
n
+
[ x 2 x 1 1
[ [ x 2 x 2 √ √ x x 2 + + x 3 x 3 x 3 x 2 2 x 3 + ··· + + x n x n x x n−1 x n+ + x 1 x 1 x 1 x n n x 1 ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( )( 1 1 x 2 + + 2 2 x 3 + ··· + x n + x 1 ≥ (x 1 + x 2 + ··· + x n−1 + x n )
+ ···
√ √
√ √
√ √ ] ⇒ (x 2 + x 3 + ··· + x n + x 1
x x 2 2 2 x 2 x x x x x x n x 1
x n n
x 1 1
x 3 3
x 2 2
x 3 3
x
n−1
··· + x x 2 2 n−1 + x x 2 2 n ≥ (x 1 + x 2 + ··· + x n−1 + x n ) 2 2 2
··· + + n−1 n n x 1≥ (x 1 + x 2 + ··· + x n−1 + x n ) ) . Divide both sides by x 1 + x 2 + ··· + x n−1 + x n ≥ 0 to obtain
x n + +
···
≥ (x 1 + x 2 + ··· + x n−1 + x n
x x
x n n
x 1 1
x 2 1 + x 2 2 + ··· + x 2 n−1 + x n 2 .
x 2 x 3 x n x 1 ≥ x 1 + x 2 + ··· + x n−1 + x n
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