The following inequality is known as Titu's lemma , Bergström's inequality , Engel's form or Sedrakyan's inequality , respectively, referring to the article About the applications of one useful inequality of Nairi Sedrakyan published in 1997,[1] to the book Problem-solving strategies of Arthur Engel published in 1998 and to the book Mathematical Olympiad Treasures of Titu Andreescu published in 2003.[2] [3]
It is a direct consequence of Cauchy–Bunyakovsky–Schwarz inequality . Nevertheless, in his article (1997) Sedrakyan has noticed that written in this form this inequality can be used as a mathematical proof technique and it has very useful new applications. In the book Algebraic Inequalities (Sedrakyan) are provided several generalizations of this inequality.[4]
Statement of the inequality
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For any reals
a
1
,
a
2
,
a
3
,
…
,
a
n
{\displaystyle a_{1},a_{2},a_{3},\ldots ,a_{n}}
and positive reals
b
1
,
b
2
,
b
3
,
…
,
b
n
,
{\displaystyle b_{1},b_{2},b_{3},\ldots ,b_{n},}
we have
a
1
2
b
1
+
a
2
2
b
2
+
⋯
+
a
n
2
b
n
≥
(
a
1
+
a
2
+
⋯
+
a
n
)
2
b
1
+
b
2
+
⋯
+
b
n
.
{\displaystyle {\frac {a_{1}^{2}}{b_{1}}}+{\frac {a_{2}^{2}}{b_{2}}}+\cdots +{\frac {a_{n}^{2}}{b_{n}}}\geq {\frac {\left(a_{1}+a_{2}+\cdots +a_{n}\right)^{2}}{b_{1}+b_{2}+\cdots +b_{n}}}.}
(Nairi Sedrakyan (1997), Arthur Engel (1998), Titu Andreescu (2003))
Probabilistic statement
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Similarly to the Cauchy–Schwarz inequality , one can generalize Sedrakyan's inequality to random variable .
In this formulation let
X
{\displaystyle X}
be a real random variable, and let
Y
{\displaystyle Y}
be a positive random variable. X and Y need not be independent, but we assume
E
[
|
X
|
]
{\displaystyle E[|X|]}
and
E
[
Y
]
{\displaystyle E[Y]}
are both defined.
Then
E
[
X
2
/
Y
]
≥
E
[
|
X
|
]
2
/
E
[
Y
]
≥
E
[
X
]
2
/
E
[
Y
]
.
{\displaystyle \operatorname {E} [X^{2}/Y]\geq \operatorname {E} [|X|]^{2}/\operatorname {E} [Y]\geq \operatorname {E} [X]^{2}/\operatorname {E} [Y].}
Direct applications
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Example 1. Nesbitt's inequality .
For positive real numbers
a
,
b
,
c
:
{\displaystyle a,b,c:}
a
b
+
c
+
b
a
+
c
+
c
a
+
b
≥
3
2
.
{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}
Example 2. International Mathematical Olympiad (IMO) 1995.
For positive real numbers
a
,
b
,
c
{\displaystyle a,b,c}
, where
a
b
c
=
1
{\displaystyle abc=1}
we have that
1
a
3
(
b
+
c
)
+
1
b
3
(
a
+
c
)
+
1
c
3
(
a
+
b
)
≥
3
2
.
{\displaystyle {\frac {1}{a^{3}(b+c)}}+{\frac {1}{b^{3}(a+c)}}+{\frac {1}{c^{3}(a+b)}}\geq {\frac {3}{2}}.}
Example 3.
For positive real numbers
a
,
b
{\displaystyle a,b}
we have that
8
(
a
4
+
b
4
)
≥
(
a
+
b
)
4
.
{\displaystyle 8(a^{4}+b^{4})\geq (a+b)^{4}.}
Example 4.
For positive real numbers
a
,
b
,
c
{\displaystyle a,b,c}
we have that
1
a
+
b
+
1
b
+
c
+
1
a
+
c
≥
9
2
(
a
+
b
+
c
)
.
{\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {9}{2(a+b+c)}}.}
Example 1.
Proof : Use
n
=
3
,
{\displaystyle n=3,}
(
a
1
,
a
2
,
a
3
)
:=
(
a
,
b
,
c
)
,
{\displaystyle \left(a_{1},a_{2},a_{3}\right):=(a,b,c),}
and
(
b
1
,
b
2
,
b
3
)
:=
(
a
(
b
+
c
)
,
b
(
c
+
a
)
,
c
(
a
+
b
)
)
{\displaystyle \left(b_{1},b_{2},b_{3}\right):=(a(b+c),b(c+a),c(a+b))}
to conclude:
a
2
a
(
b
+
c
)
+
b
2
b
(
c
+
a
)
+
c
2
c
(
a
+
b
)
≥
(
a
+
b
+
c
)
2
a
(
b
+
c
)
+
b
(
c
+
a
)
+
c
(
a
+
b
)
=
a
2
+
b
2
+
c
2
+
2
(
a
b
+
b
c
+
c
a
)
2
(
a
b
+
b
c
+
c
a
)
=
a
2
+
b
2
+
c
2
2
(
a
b
+
b
c
+
c
a
)
+
1
≥
1
2
(
1
)
+
1
=
3
2
.
◼
{\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {(a+b+c)^{2}}{a(b+c)+b(c+a)+c(a+b)}}={\frac {a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}{2(ab+bc+ca)}}={\frac {a^{2}+b^{2}+c^{2}}{2(ab+bc+ca)}}+1\geq {\frac {1}{2}}(1)+1={\frac {3}{2}}.\blacksquare }
Example 2.
We have that
(
1
a
)
2
a
(
b
+
c
)
+
(
1
b
)
2
b
(
a
+
c
)
+
(
1
c
)
2
c
(
a
+
b
)
≥
(
1
a
+
1
b
+
1
c
)
2
2
(
a
b
+
b
c
+
a
c
)
=
a
b
+
b
c
+
a
c
2
a
2
b
2
c
2
≥
3
a
2
b
2
c
2
3
2
a
2
b
2
c
2
=
3
2
.
{\displaystyle {\frac {{\Big (}{\frac {1}{a}}{\Big )}^{2}}{a(b+c)}}+{\frac {{\Big (}{\frac {1}{b}}{\Big )}^{2}}{b(a+c)}}+{\frac {{\Big (}{\frac {1}{c}}{\Big )}^{2}}{c(a+b)}}\geq {\frac {{\Big (}{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}{\Big )}^{2}}{2(ab+bc+ac)}}={\frac {ab+bc+ac}{2a^{2}b^{2}c^{2}}}\geq {\frac {3{\sqrt[{3}]{a^{2}b^{2}c^{2}}}}{2a^{2}b^{2}c^{2}}}={\frac {3}{2}}.}
Example 3.
We have
a
2
1
+
b
2
1
≥
(
a
+
b
)
2
2
{\displaystyle {\frac {a^{2}}{1}}+{\frac {b^{2}}{1}}\geq {\frac {(a+b)^{2}}{2}}}
so that
a
4
+
b
4
=
(
a
2
)
2
1
+
(
b
2
)
2
1
≥
(
a
2
+
b
2
)
2
2
≥
(
(
a
+
b
)
2
2
)
2
2
=
(
a
+
b
)
4
8
.
{\displaystyle a^{4}+b^{4}={\frac {\left(a^{2}\right)^{2}}{1}}+{\frac {\left(b^{2}\right)^{2}}{1}}\geq {\frac {\left(a^{2}+b^{2}\right)^{2}}{2}}\geq {\frac {\left({\frac {(a+b)^{2}}{2}}\right)^{2}}{2}}={\frac {(a+b)^{4}}{8}}.}
Example 4.
We have that
1
a
+
b
+
1
b
+
c
+
1
a
+
c
≥
(
1
+
1
+
1
)
2
2
(
a
+
b
+
c
)
=
9
2
(
a
+
b
+
c
)
.
{\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {(1+1+1)^{2}}{2(a+b+c)}}={\frac {9}{2(a+b+c)}}.}
References
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