Teorema: Tegul
[
x
,
y
]
{\displaystyle [x,y]}
vektorių skaliarinė sandauga, tai su vektoriais x ir y , priklausantiems šiai erdvei, galioja nelygybė:
[
x
,
x
]
[
y
,
y
]
≥
|
[
x
,
y
]
|
2
{\displaystyle [x,x][y,y]\geq |[x,y]|^{2}}
be to, lygybė galioja tada ir tik tada kai vektoriai x ir y yra tiesiškai priklausomi.
Teorema: Bet kokioms tolydžioms ir integruojamoms intervale
[
a
,
b
]
{\displaystyle [a,b]}
funkcijoms
f
:
R
→
R
{\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }
g
:
R
→
R
{\displaystyle g:\mathbb {R} \rightarrow \mathbb {R} }
∫
a
b
(
f
(
x
)
)
2
d
x
∫
a
b
(
g
(
x
)
)
2
d
x
≥
(
∫
a
b
f
(
x
)
g
(
x
)
d
x
)
2
{\displaystyle \int \limits _{a}^{b}\left(f(x)\right)^{2}dx\int \limits _{a}^{b}\left(g(x)\right)^{2}dx\geq \left(\int \limits _{a}^{b}f(x)g(x)dx\right)^{2}}
Lygybė galioja tada ir tik tada, kai egzistuoja toks
c
∈
R
{\displaystyle c\in \mathbb {R} }
f
(
x
)
=
c
g
(
x
)
{\displaystyle f(x)=cg(x)}
Teorema: Tarkime
(
x
n
)
n
≥
1
{\displaystyle (x_{n})_{n\geq 1}}
(
y
n
)
n
≥
1
{\displaystyle (y_{n})_{n\geq 1}}
sekos , tuomet galioja nelygybė:
(
∑
i
=
1
n
x
i
2
)
(
∑
i
=
1
n
y
i
2
)
≥
(
∑
i
=
1
n
x
i
y
i
)
2
{\displaystyle \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)\geq \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}}
Lygybės atvejis pasiekiamas tada ir tik tada, kai:
x
1
y
1
=
x
2
y
2
=
.
.
.
=
x
n
y
n
.
{\displaystyle {\frac {x_{1}}{y_{1}}}={\frac {x_{2}}{y_{2}}}=...={\frac {x_{n}}{y_{n}}}.}
Nelygybė lengvai įrodoma pasinaudojus Lagranžo tapatybe:
(
∑
i
=
1
n
x
i
2
)
(
∑
i
=
1
n
y
i
2
)
−
(
∑
i
=
1
n
y
i
x
i
)
2
=
∑
1
≤
i
<
j
≤
n
(
x
i
y
j
−
x
j
y
i
)
2
≥
0
⇒
{\displaystyle \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)-\left(\sum _{i=1}^{n}y_{i}x_{i}\right)^{2}=\sum _{1\leq i<j\leq n}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\geq 0\Rightarrow }
(
∑
i
=
1
n
x
i
2
)
(
∑
i
=
1
n
y
i
2
)
≥
(
∑
i
=
1
n
x
i
y
i
)
2
▴
{\displaystyle \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)\geq \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}\blacktriangle }
Tarkime
(
a
n
)
n
≥
1
{\displaystyle (a_{n})_{n\geq 1}}
(
b
n
)
n
≥
1
{\displaystyle (b_{n})_{n\geq 1}}
kvadratinė funkcija
f
(
x
)
=
∑
i
=
1
n
(
a
i
x
−
b
i
)
2
{\displaystyle f(x)=\sum _{i=1}^{n}\left(a_{i}x-b_{i}\right)^{2}}
x
∈
R
{\displaystyle x\in \mathbb {R} }
D turi būti mažesnis arba lygus už nulį. Taigi:
f
(
x
)
=
∑
i
=
1
n
(
a
i
x
−
b
i
)
2
=
x
2
∑
i
=
1
n
a
i
2
−
2
x
∑
i
=
1
n
a
i
b
i
+
∑
i
=
1
n
b
i
2
≥
0
{\displaystyle f(x)=\sum _{i=1}^{n}\left(a_{i}x-b_{i}\right)^{2}=x^{2}\sum _{i=1}^{n}a_{i}^{2}-2x\sum _{i=1}^{n}a_{i}b_{i}+\sum _{i=1}^{n}b_{i}^{2}\geq 0}
D
≤
0
⇒
{\displaystyle D\leq 0\Rightarrow }
D
=
4
(
∑
i
=
1
n
a
i
b
i
)
2
−
4
(
∑
i
=
1
n
a
i
2
)
(
∑
i
=
1
n
b
i
2
)
≤
0
⇒
{\displaystyle D=4\left(\sum _{i=1}^{n}a_{i}b_{i}\right)^{2}-4\left(\sum _{i=1}^{n}a_{i}^{2}\right)\left(\sum _{i=1}^{n}b_{i}^{2}\right)\leq 0\Rightarrow }
(
∑
i
=
1
n
a
i
2
)
(
∑
i
=
1
n
b
i
2
)
≥
(
∑
i
=
1
n
a
i
b
i
)
2
▴
{\displaystyle \left(\sum _{i=1}^{n}a_{i}^{2}\right)\left(\sum _{i=1}^{n}b_{i}^{2}\right)\geq \left(\sum _{i=1}^{n}a_{i}b_{i}\right)^{2}\blacktriangle }
Pažymėkime
A
=
∑
i
=
1
n
a
i
2
{\displaystyle A={\sqrt {\sum _{i=1}^{n}a_{i}^{2}}}}
B
=
∑
i
=
1
n
b
i
2
{\displaystyle B={\sqrt {\sum _{i=1}^{n}b_{i}^{2}}}}
aritmetinio – geometrinio vidurkio (AM
≥
{\displaystyle \geq }
) nelygybę turime:
∑
i
=
1
n
a
i
b
i
A
B
≤
1
2
∑
i
=
1
n
(
a
i
2
A
2
+
b
i
2
B
2
)
=
1
⇒
∑
i
=
1
n
a
i
b
i
≤
A
B
{\displaystyle \sum _{i=1}^{n}{\frac {a_{i}b_{i}}{AB}}\leq {\frac {1}{2}}\sum _{i=1}^{n}\left({\frac {a_{i}^{2}}{A^{2}}}+{\frac {b_{i}^{2}}{B^{2}}}\right)=1\Rightarrow \sum _{i=1}^{n}a_{i}b_{i}\leq AB}
▴
{\displaystyle \blacktriangle }
Akivaizdu, kad funkcija
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
f
″
(
x
)
>
0
{\displaystyle f''(x)>0}
∑
i
=
1
n
w
i
x
i
2
≥
(
∑
i
=
1
n
w
i
x
i
)
2
{\displaystyle \sum _{i=1}^{n}w_{i}x_{i}^{2}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}\right)^{2}}
w
i
{\displaystyle w_{i}}
∑
i
=
1
n
w
i
=
1
;
{\displaystyle \sum _{i=1}^{n}w_{i}=1;}
Dabar kiekvienam
b
i
≠
0
{\displaystyle b_{i}\neq 0}
i
=
1
,
2
,
3
,
.
.
.
,
n
{\displaystyle i=1,2,3,...,n}
x
i
=
a
i
b
i
{\displaystyle x_{i}={\frac {a_{i}}{b_{i}}}}
w
i
=
b
i
2
∑
i
=
1
n
b
i
2
{\displaystyle w_{i}={\frac {b_{i}^{2}}{\sum _{i=1}^{n}b_{i}^{2}}}}
▴
{\displaystyle \blacktriangle }
Tarkime
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
x
,
y
∈
R
+
{\displaystyle x,y\in \mathbb {R^{+}} }
(
a
y
−
b
x
)
2
≥
0
⇒
{\displaystyle \left(ay-bx\right)^{2}\geq 0\Rightarrow }
a
2
y
2
+
b
2
x
2
−
2
a
b
x
y
≥
0
⇒
{\displaystyle a^{2}y^{2}+b^{2}x^{2}-2abxy\geq 0\Rightarrow }
a
2
y
x
+
b
2
x
y
≥
2
a
b
⇒
{\displaystyle {\frac {a^{2}y}{x}}+{\frac {b^{2}x}{y}}\geq 2ab\Rightarrow }
a
2
y
x
+
b
2
x
y
+
a
2
+
b
2
≥
(
a
+
b
)
2
{\displaystyle {\frac {a^{2}y}{x}}+{\frac {b^{2}x}{y}}+a^{2}+b^{2}\geq \left(a+b\right)^{2}}
a
2
x
+
b
2
y
≥
(
a
+
b
)
2
x
+
y
.
{\displaystyle {\frac {a^{2}}{x}}+{\frac {b^{2}}{y}}\geq {\frac {(a+b)^{2}}{x+y}}.}
b
{\displaystyle b}
b
+
c
{\displaystyle b+c}
c
∈
R
{\displaystyle c\in \mathbb {R} }
y
{\displaystyle y}
y
+
z
{\displaystyle y+z}
z
∈
R
+
{\displaystyle z\in \mathbb {R^{+}} }
a
2
x
+
b
2
y
+
c
2
z
≥
(
a
+
b
+
c
)
2
x
+
y
+
z
{\displaystyle {\frac {a^{2}}{x}}+{\frac {b^{2}}{y}}+{\frac {c^{2}}{z}}\geq {\frac {(a+b+c)^{2}}{x+y+z}}}
n kartų, gausime apibendrintą Tito lemą, dar vadinamą Koši – Švarco nelygybe Engel formoje (matematiko Artūro Engelio garbei):
∑
i
=
1
n
a
i
2
b
i
≥
(
∑
i
=
1
n
a
i
)
2
∑
i
=
1
n
b
i
.
{\displaystyle \sum _{i=1}^{n}{\frac {a_{i}^{2}}{b_{i}}}\geq {\frac {\left(\sum _{i=1}^{n}a_{i}\right)^{2}}{\sum _{i=1}^{n}b_{i}}}.}
a
i
=
α
i
β
i
{\displaystyle a_{i}=\alpha _{i}\beta _{i}}
b
i
=
β
i
2
{\displaystyle b_{i}=\beta _{i}^{2}}
(
∑
i
=
1
n
α
i
2
)
(
∑
i
=
1
n
β
i
2
)
≥
(
∑
i
=
1
n
α
i
β
i
)
2
▴
{\displaystyle \left(\sum _{i=1}^{n}\alpha _{i}^{2}\right)\left(\sum _{i=1}^{n}\beta _{i}^{2}\right)\geq \left(\sum _{i=1}^{n}\alpha _{i}\beta _{i}\right)^{2}\blacktriangle }
![{\displaystyle [x,y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b7bd6292c6023626c6358bfd3943a031b27d663)
![{\displaystyle [x,x][y,y]\geq |[x,y]|^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1545cbd7b836dfac4e7c2171db5a0dfa86506e83)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
