Неравенството на Карамата, наречено на сръбския математик Йован Карамата, още известно като мажорационно неравенство, е теорема от елементарната алгебра.
Ако е дадена функция
f
:
I
→
R
{\displaystyle f:{\mathcal {I}}\rightarrow \mathbb {R} }
I
{\displaystyle {\mathcal {I}}}
{
x
1
,
x
2
,
…
,
x
n
}
≻
{
y
1
,
y
2
,
…
,
y
n
}
{\displaystyle \{x_{1},x_{2},\ldots ,x_{n}\}\succ \{y_{1},y_{2},\ldots ,y_{n}\}}
f
(
x
1
)
+
⋯
+
f
(
x
n
)
≥
f
(
y
1
)
+
⋯
+
f
(
y
n
)
.
{\displaystyle f(x_{1})+\cdots +f(x_{n})\geq f(y_{1})+\cdots +f(y_{n}).}
Доказателство:
Първо нека положим
c
i
=
f
(
y
i
)
−
f
(
x
i
)
y
i
−
x
i
{\displaystyle c_{i}={\frac {f(y_{i})-f(x_{i})}{y_{i}-x_{i}}}}
f
(
x
)
{\displaystyle f(x)}
x
{\displaystyle \mathbf {x} }
y
{\displaystyle \mathbf {y} }
c
i
≥
c
i
+
1
⇔
f
(
y
i
)
−
f
(
x
i
)
y
i
−
x
i
≥
f
(
y
i
+
1
)
−
f
(
x
i
+
1
)
y
i
+
1
−
x
i
+
1
.
{\displaystyle c_{i}\geq c_{i+1}\Leftrightarrow {\frac {f(y_{i})-f(x_{i})}{y_{i}-x_{i}}}\geq {\frac {f(y_{i+1})-f(x_{i+1})}{y_{i+1}-x_{i+1}}}.}
Това следва последователно от
b
−
c
a
−
c
f
(
a
)
+
a
−
b
a
−
c
f
(
c
)
≥
f
(
b
)
=
f
(
b
−
c
a
−
c
a
+
a
−
b
a
−
c
c
)
⇔
f
(
a
)
−
f
(
c
)
a
−
c
≥
f
(
b
)
−
f
(
c
)
b
−
c
{\displaystyle {\frac {b-c}{a-c}}f(a)+{\frac {a-b}{a-c}}f(c)\geq f(b)=f({\frac {b-c}{a-c}}a+{\frac {a-b}{a-c}}c)\Leftrightarrow {\frac {f(a)-f(c)}{a-c}}\geq {\frac {f(b)-f(c)}{b-c}}}
a
≥
b
{\displaystyle a\geq b}
x
1
≥
x
2
≥
…
≥
x
n
{\displaystyle x_{1}\geq x_{2}\geq \ldots \geq x_{n}}
y
1
≥
y
2
≥
…
≥
y
n
{\displaystyle y_{1}\geq y_{2}\geq \ldots \geq y_{n}}
f
(
y
i
)
−
f
(
x
i
)
y
i
−
x
i
≥
f
(
y
i
)
−
f
(
x
i
+
1
)
y
i
−
x
i
+
1
≥
f
(
y
i
+
1
)
−
f
(
x
i
+
1
)
y
i
+
1
−
x
i
+
1
.
{\displaystyle {\frac {f(y_{i})-f(x_{i})}{y_{i}-x_{i}}}\geq {\frac {f(y_{i})-f(x_{i+1})}{y_{i}-x_{i+1}}}\geq {\frac {f(y_{i+1})-f(x_{i+1})}{y_{i+1}-x_{i+1}}}.}
Полагаме
A
k
=
x
1
+
…
+
x
k
{\displaystyle A_{k}=x_{1}+\ldots +x_{k}}
B
k
=
y
1
+
…
+
y
k
{\displaystyle B_{k}=y_{1}+\ldots +y_{k}}
k
=
1
,
2
,
…
,
n
{\displaystyle k=1,2,\ldots ,n}
A
k
≥
B
k
{\displaystyle A_{k}\geq B_{k}}
k
=
1
,
2
,
…
,
n
−
1
{\displaystyle k=1,2,\ldots ,n-1}
A
n
=
B
n
{\displaystyle A_{n}=B_{n}}
В такъв случай
∑
i
=
1
n
(
f
(
x
i
)
−
f
(
y
i
)
)
=
∑
i
=
1
n
c
i
(
x
i
−
y
i
)
=
{\displaystyle \sum _{i=1}^{n}(f(x_{i})-f(y_{i}))=\sum _{i=1}^{n}c_{i}(x_{i}-y_{i})=}
∑
i
=
1
n
(
c
i
−
c
i
+
1
)
(
x
1
+
x
2
+
.
.
.
+
x
i
)
−
∑
i
=
1
n
(
c
i
−
c
i
+
1
)
(
y
1
+
y
2
+
.
.
.
+
y
i
)
=
{\displaystyle \sum _{i=1}^{n}(c_{i}-c_{i+1})(x_{1}+x_{2}+...+x_{i})-\sum _{i=1}^{n}(c_{i}-c_{i+1})(y_{1}+y_{2}+...+y_{i})=}
=
∑
i
=
1
n
−
1
(
c
i
−
c
i
+
1
)
(
A
i
−
B
i
)
+
c
n
(
A
n
−
B
n
)
{\displaystyle =\sum _{i=1}^{n-1}(c_{i}-c_{i+1})(A_{i}-B_{i})+c_{n}(A_{n}-B_{n})}
Ако вместо
(
y
1
,
y
2
,
…
,
y
n
)
{\displaystyle (y_{1},y_{2},\ldots ,y_{n})}
(
x
1
+
⋯
+
x
n
n
,
⋯
,
x
1
+
⋯
+
x
n
n
)
{\displaystyle \left({\frac {x_{1}+\cdots +x_{n}}{n}},\cdots ,{\frac {x_{1}+\cdots +x_{n}}{n}}\right)}
