๐ย ยท ๐ ๐ ๐ ๐ ๐ ๐= โค ๐ ๐ ๐= ๐ ๐ ๐= (using lagrange...
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๐
๐
๐๐๐๐
๐
๐=๐
๐
โค ๐๐๐
๐
๐=๐
๐๐๐
๐
๐=๐
(
Using Lagrange Identity
Algebraic method
Using mathematical
induction
Using arithmetic-geometric
means
Geometric proof
Using convex function and
Jensenโs inequality
Some methods of proof
For all complex numbers ๐1, ๐2, โฆ , ๐๐ and ๐1, ๐2, โฆ , ๐๐
there is an inequality
๐๐๐ ๐
๐
๐=1
2
โค ๐๐2
๐
๐=1
๐๐2
๐
๐=1
.
Consider sequences of complex numbers {๐ข๐}๐=1๐ and ๐ฃ๐ ๐=1
๐ , where ๐ข๐ = ๐2๐โ1 + ๐๐2๐ , ๐ฃ๐ = ๐2๐โ1 + ๐๐2๐ .
Notice, that
๐ข๐๐ฃ ๐
๐
๐=1
2
= Re ๐ข๐๐ฃ ๐
๐
๐=1
2
+ Im ๐ข๐๐ฃ ๐
๐
๐=1
2
=
= Re ๐2๐โ1 + ๐๐2๐ ๐2๐โ1 โ ๐๐2๐
๐
๐=1
2
+ Im ๐2๐โ1 + ๐๐2๐ ๐2๐โ1 โ ๐๐2๐
๐
๐=1
2
=
= ๐2๐โ1๐2๐โ1 + ๐2๐๐2๐
๐
๐=1
2
+ ๐2๐๐2๐โ1 โ ๐2๐โ1๐2๐
๐
๐=1
2
= ๐๐๐๐
2๐
๐=1
2
+
+ ๐2๐๐2๐โ1 โ ๐2๐โ1๐2๐
๐
๐=1
2
.
๐ข๐2
๐
๐=1
๐ฃ๐2
๐
๐=1
= (๐2๐โ12 + ๐2๐
2 )
๐
๐=1
(๐2๐โ12 + ๐2๐
2 )
๐
๐=1
= ๐๐2
2๐
๐=1
๐๐2
2๐
๐=1
.
Hence, the Cauchy-Bunyakovsky inequality has the form:
๐๐2
2๐
๐=1
๐๐2
2๐
๐=1
โ ๐2๐๐2๐โ1 โ ๐2๐โ1๐2๐
๐
๐=1
2
โฅ ๐๐๐๐
2๐
๐=1
2
๐๐๐
๐๐
๐=๐
๐๐๐
๐๐
๐=๐
โ ๐๐๐๐๐๐โ๐ โ ๐๐๐โ๐๐๐๐
๐
๐=๐
๐
โฅ ๐๐๐๐
๐๐
๐=๐
๐
๐
If ๐ข1, โฆ , ๐ข๐ and ๐ฃ1, โฆ , ๐ฃ๐ are quaternion, then the
Cauchy-Bunyakovsky inequality has a form:
๐ข๐ก๐ฃ ๐ก
๐
๐ก=1
2
โค ๐ข๐ก2
๐
๐ก=1
๐ฃ๐ก2
๐
๐ก=1
.
Denote ๐ข๐ก = ๐4๐กโ3 + ๐๐4๐กโ2 + ๐๐4๐กโ1 + ๐๐4๐ก , ๐ฃ๐ก = ๐4๐กโ3 + ๐๐4๐กโ2 + ๐๐4๐กโ1 + ๐๐4๐ก.
Then ๐ข๐ก2 = ๐4๐กโ3
2 + ๐4๐กโ22 + ๐4๐กโ1
2 + ๐4๐ก2 ,
๐ฃ๐ก2 = ๐4๐กโ3
2 + ๐4๐กโ22 + ๐4๐กโ1
2 + ๐4๐ก2 ,
๐ข๐ก2
๐
๐ก=1
= ๐4๐กโ32 + ๐4๐กโ2
2 + ๐4๐กโ12 + ๐4๐ก
2
๐
๐ก=1
= ๐๐ก2
4๐
๐ก=1
,
๐ฃ๐ก2
๐
๐ก=1
= ๐4๐กโ32 + ๐4๐กโ2
2 + ๐4๐กโ12 + ๐4๐ก
2
๐
๐ก=1
= ๐๐ก2
4๐
๐ก=1
.
๐๐๐๐
๐๐
๐=๐
๐
+ ๐๐๐๐๐๐โ๐ โ ๐๐๐โ๐๐๐๐
๐๐
๐=๐
๐
+
+ ๐๐๐โ๐๐๐๐โ๐ โ ๐๐๐โ๐๐๐๐โ๐ + ๐๐๐โ๐๐๐๐ โ ๐๐๐๐๐๐โ๐
๐
๐=๐
๐
+
+ ๐๐๐โ๐๐๐๐โ๐ โ ๐๐๐โ๐๐๐๐โ๐ + ๐๐๐๐๐๐โ๐ โ ๐๐๐โ๐๐๐๐
๐
๐=๐
๐
โค ๐๐๐
๐๐
๐=๐
โ ๐๐๐
๐๐
๐=๐
.
๐๐
๐ = ๐ + ๐๐ + ๐๐ + ๐๐, ๐2 = ๐2 = ๐2 = โ1