𝑛 · 𝑛 𝑛 𝒌 𝒌 𝒏 𝒌= ≤ 𝒌 𝒏 𝒌= 𝒌 𝒏 𝒌= (Using Lagrange Identity Algebraic method Using mathematical

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𝑛

𝑛

𝒂𝒌𝒃𝒌

𝒏

𝒌=𝟏

𝟐

≤ 𝒂𝒌𝟐

𝒏

𝒌=𝟏

𝒃𝒌𝟐

𝒏

𝒌=𝟏

(

Using Lagrange Identity

Algebraic method

Using mathematical

induction

Using arithmetic-geometric

means

Geometric proof

Using convex function and

Jensen’s inequality

Some methods of proof

For all complex numbers 𝑎1, 𝑎2, … , 𝑎𝑛 and 𝑏1, 𝑏2, … , 𝑏𝑛

there is an inequality

𝑎𝑘𝑏 𝑘

𝑛

𝑘=1

2

≤ 𝑎𝑘2

𝑛

𝑘=1

𝑏𝑘2

𝑛

𝑘=1

.

Consider sequences of complex numbers {𝑢𝑘}𝑘=1𝑛 and 𝑣𝑘 𝑘=1

𝑛 , where 𝑢𝑘 = 𝑎2𝑘−1 + 𝑖𝑎2𝑘 , 𝑣𝑘 = 𝑏2𝑘−1 + 𝑖𝑏2𝑘 .

Notice, that

𝑢𝑘𝑣 𝑘

𝑛

𝑘=1

2

= Re 𝑢𝑘𝑣 𝑘

𝑛

𝑘=1

2

+ Im 𝑢𝑘𝑣 𝑘

𝑛

𝑘=1

2

=

= Re 𝑎2𝑘−1 + 𝑖𝑎2𝑘 𝑏2𝑘−1 − 𝑖𝑏2𝑘

𝑛

𝑘=1

2

+ Im 𝑎2𝑘−1 + 𝑖𝑎2𝑘 𝑏2𝑘−1 − 𝑖𝑏2𝑘

𝑛

𝑘=1

2

=

= 𝑎2𝑘−1𝑏2𝑘−1 + 𝑎2𝑘𝑏2𝑘

𝑛

𝑘=1

2

+ 𝑎2𝑘𝑏2𝑘−1 − 𝑎2𝑘−1𝑏2𝑘

𝑛

𝑘=1

2

= 𝑎𝑘𝑏𝑘

2𝑛

𝑘=1

2

+

+ 𝑎2𝑘𝑏2𝑘−1 − 𝑎2𝑘−1𝑏2𝑘

𝑛

𝑘=1

2

.

𝑢𝑘2

𝑛

𝑘=1

𝑣𝑘2

𝑛

𝑘=1

= (𝑎2𝑘−12 + 𝑎2𝑘

2 )

𝑛

𝑘=1

(𝑏2𝑘−12 + 𝑏2𝑘

2 )

𝑛

𝑘=1

= 𝑎𝑘2

2𝑛

𝑘=1

𝑏𝑘2

2𝑛

𝑘=1

.

Hence, the Cauchy-Bunyakovsky inequality has the form:

𝑎𝑘2

2𝑛

𝑘=1

𝑏𝑘2

2𝑛

𝑘=1

− 𝑎2𝑘𝑏2𝑘−1 − 𝑎2𝑘−1𝑏2𝑘

𝑛

𝑘=1

2

≥ 𝑎𝑘𝑏𝑘

2𝑛

𝑘=1

2

𝒂𝒌𝟐

𝟐𝒏

𝒌=𝟏

𝒃𝒌𝟐

𝟐𝒏

𝒌=𝟏

− 𝒂𝟐𝒌𝒃𝟐𝒌−𝟏 − 𝒂𝟐𝒌−𝟏𝒃𝟐𝒌

𝒏

𝒌=𝟏

𝟐

≥ 𝒂𝒌𝒃𝒌

𝟐𝒏

𝒌=𝟏

𝟐

𝒏

If 𝑢1, … , 𝑢𝑛 and 𝑣1, … , 𝑣𝑛 are quaternion, then the

Cauchy-Bunyakovsky inequality has a form:

𝑢𝑡𝑣 𝑡

𝑛

𝑡=1

2

≤ 𝑢𝑡2

𝑛

𝑡=1

𝑣𝑡2

𝑛

𝑡=1

.

Denote 𝑢𝑡 = 𝑎4𝑡−3 + 𝑖𝑎4𝑡−2 + 𝑗𝑎4𝑡−1 + 𝑘𝑎4𝑡 , 𝑣𝑡 = 𝑏4𝑡−3 + 𝑖𝑏4𝑡−2 + 𝑗𝑏4𝑡−1 + 𝑘𝑏4𝑡.

Then 𝑢𝑡2 = 𝑎4𝑡−3

2 + 𝑎4𝑡−22 + 𝑎4𝑡−1

2 + 𝑎4𝑡2 ,

𝑣𝑡2 = 𝑏4𝑡−3

2 + 𝑏4𝑡−22 + 𝑏4𝑡−1

2 + 𝑏4𝑡2 ,

𝑢𝑡2

𝑛

𝑡=1

= 𝑎4𝑡−32 + 𝑎4𝑡−2

2 + 𝑎4𝑡−12 + 𝑎4𝑡

2

𝑛

𝑡=1

= 𝑎𝑡2

4𝑛

𝑡=1

,

𝑣𝑡2

𝑛

𝑡=1

= 𝑏4𝑡−32 + 𝑏4𝑡−2

2 + 𝑏4𝑡−12 + 𝑏4𝑡

2

𝑛

𝑡=1

= 𝑏𝑡2

4𝑛

𝑡=1

.

𝒂𝒕𝒃𝒕

𝟒𝒏

𝒕=𝟏

𝟐

+ 𝒂𝟐𝒕𝒃𝟐𝒕−𝟏 − 𝒂𝟐𝒕−𝟏𝒃𝟐𝒕

𝟐𝒏

𝒕=𝟏

𝟐

+

+ 𝒂𝟒𝒕−𝟏𝒃𝟒𝒕−𝟑 − 𝒂𝟒𝒕−𝟑𝒃𝟒𝒕−𝟏 + 𝒂𝟒𝒕−𝟐𝒃𝟒𝒕 − 𝒂𝟒𝒕𝒃𝟒𝒕−𝟐

𝒏

𝒕=𝟏

𝟐

+

+ 𝒂𝟒𝒕−𝟏𝒃𝟒𝒕−𝟐 − 𝒂𝟒𝒕−𝟐𝒃𝟒𝒕−𝟏 + 𝒂𝟒𝒕𝒃𝟒𝒕−𝟑 − 𝒂𝟒𝒕−𝟑𝒃𝟒𝒕

𝒏

𝒕=𝟏

𝟐

≤ 𝒂𝒕𝟐

𝟒𝒏

𝒕=𝟏

⋅ 𝒃𝒕𝟐

𝟒𝒏

𝒕=𝟏

.

𝟒𝒏

𝑞 = 𝑎 + 𝑖𝑏 + 𝑗𝑐 + 𝑘𝑑, 𝑖2 = 𝑗2 = 𝑘2 = −1