The Real And Complex Number Systems

and
( n∑
) ( n∑
)
a k b k a k b k = ∑
a k b k a j b j =
k=1
k=1
1≤k,j≤n
So,
( n∑
k=1
a k b k
) 2
=
=
=
( n∑
k=1
( n∑
k=1
( n∑
k=1
a 2 k
a 2 k
a 2 k
) ( n∑
k=1
) ( n∑
k=1
) ( n∑
k=1
b 2 k
b 2 k
b 2 k
)
n∑
k=1
a 2 kb 2 k + ∑ k≠j
+ ∑ a k b k a j b j − ∑ a 2 kb 2 j
k≠j
k≠j
)
∑
+ 2 a k b k a j b j −
∑
)
−
1≤k| ≤ ‖x‖ ‖y‖ by Remark (1).
1.24 Prove that for arbitrary real a k , b k , c k we have
( n∑
) 4 ( n∑
) ( n∑
) 2 ( n∑
)
a k b k c k ≤ a 4 k b 2 k c 4 k .
k=1
k=1 k=1 k=1
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