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π
π
ππππ
π
π=π
π
β€ πππ
π
π=π
πππ
π
π=π
(
Using Lagrange Identity
Algebraic method
Using mathematical
induction
Using arithmetic-geometric
means
Geometric proof
Using convex function and
Jensenβs inequality
Some methods of proof
For all complex numbers π1, π2, β¦ , ππ and π1, π2, β¦ , ππ
there is an inequality
πππ π
π
π=1
2
β€ ππ2
π
π=1
ππ2
π
π=1
.
Consider sequences of complex numbers {π’π}π=1π and π£π π=1
π , where π’π = π2πβ1 + ππ2π , π£π = π2πβ1 + ππ2π .
Notice, that
π’ππ£ π
π
π=1
2
= Re π’ππ£ π
π
π=1
2
+ Im π’ππ£ π
π
π=1
2
=
= Re π2πβ1 + ππ2π π2πβ1 β ππ2π
π
π=1
2
+ Im π2πβ1 + ππ2π π2πβ1 β ππ2π
π
π=1
2
=
= π2πβ1π2πβ1 + π2ππ2π
π
π=1
2
+ π2ππ2πβ1 β π2πβ1π2π
π
π=1
2
= ππππ
2π
π=1
2
+
+ π2ππ2πβ1 β π2πβ1π2π
π
π=1
2
.
π’π2
π
π=1
π£π2
π
π=1
= (π2πβ12 + π2π
2 )
π
π=1
(π2πβ12 + π2π
2 )
π
π=1
= ππ2
2π
π=1
ππ2
2π
π=1
.
Hence, the Cauchy-Bunyakovsky inequality has the form:
ππ2
2π
π=1
ππ2
2π
π=1
β π2ππ2πβ1 β π2πβ1π2π
π
π=1
2
β₯ ππππ
2π
π=1
2
πππ
ππ
π=π
πππ
ππ
π=π
β ππππππβπ β πππβππππ
π
π=π
π
β₯ ππππ
ππ
π=π
π
π
If π’1, β¦ , π’π and π£1, β¦ , π£π are quaternion, then the
Cauchy-Bunyakovsky inequality has a form:
π’π‘π£ π‘
π
π‘=1
2
β€ π’π‘2
π
π‘=1
π£π‘2
π
π‘=1
.
Denote π’π‘ = π4π‘β3 + ππ4π‘β2 + ππ4π‘β1 + ππ4π‘ , π£π‘ = π4π‘β3 + ππ4π‘β2 + ππ4π‘β1 + ππ4π‘.
Then π’π‘2 = π4π‘β3
2 + π4π‘β22 + π4π‘β1
2 + π4π‘2 ,
π£π‘2 = π4π‘β3
2 + π4π‘β22 + π4π‘β1
2 + π4π‘2 ,
π’π‘2
π
π‘=1
= π4π‘β32 + π4π‘β2
2 + π4π‘β12 + π4π‘
2
π
π‘=1
= ππ‘2
4π
π‘=1
,
π£π‘2
π
π‘=1
= π4π‘β32 + π4π‘β2
2 + π4π‘β12 + π4π‘
2
π
π‘=1
= ππ‘2
4π
π‘=1
.
ππππ
ππ
π=π
π
+ ππππππβπ β πππβππππ
ππ
π=π
π
+
+ πππβππππβπ β πππβππππβπ + πππβππππ β ππππππβπ
π
π=π
π
+
+ πππβππππβπ β πππβππππβπ + ππππππβπ β πππβππππ
π
π=π
π
β€ πππ
ππ
π=π
β πππ
ππ
π=π
.
ππ
π = π + ππ + ππ + ππ, π2 = π2 = π2 = β1