Titu's lemma explained

In mathematics, 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 proof technique and it has very useful new applications. In the book Algebraic Inequalities (Sedrakyan) several generalizations of this inequality are provided.[4]

Statement of the inequality

For any real numbers

a1,a2,a3,\ldots,an

and positive reals

b1,b2,b3,\ldots,bn,

we have
2
a
1
b1

+

2
a
2
b2

++

2
a
n
bn

\geq

\left(a+a2++
2
a
n\right)
1
b1+b2+ … +bn

.

(Nairi Sedrakyan (1997), Arthur Engel (1998), Titu Andreescu (2003))

Probabilistic statement

Similarly to the Cauchy–Schwarz inequality, one can generalize Sedrakyan's inequality to random variables.In this formulation let

X

be a real random variable, and let

Y

be a positive random variable. X and Y need not be independent, but we assume

E[|X|]

and

E[Y]

are both defined.Then\operatorname[X^2/Y] \ge \operatorname[|X|]^2 / \operatorname[Y] \ge \operatorname[X]^2 / \operatorname[Y].

Direct applications

Example 1. Nesbitt's inequality.

For positive real numbers

a,b,c:

a
b+c

+

b
a+c

+

c
a+b

\geq

3
2

.

Example 2. International Mathematical Olympiad (IMO) 1995.

For positive real numbers

a,b,c

, where

abc=1

we have that
1+
a3(b+c)
1+
b3(a+c)
1
c3(a+b)

\geq

3
2

.

Example 3.

For positive real numbers

a,b

we have that

8(a4+b4)\geq(a+b)4.

Example 4.

For positive real numbers

a,b,c

we have that
1+
a+b
1+
b+c
1
a+c

\geq

9
2(a+b+c)

.

Proofs

Example 1.

Proof: Use

n=3,

\left(a1,a2,a3\right):=(a,b,c),

and

\left(b1,b2,b3\right):=(a(b+c),b(c+a),c(a+b))

to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare

Example 2.

We have that

(1)2
a
a(b+c)

+

(1)2
b
b(a+c)

+

(1)2
c
c(a+b)

\geq

(1
+1
b
+1
c
)2
a
2(ab+bc+ac)

=

ab+bc+ac
2a2b2c2

\geq

3\sqrt[3]{a2b2c2
} = \frac.

Example 3.

We have

a2
1

+

b2
1

\geq

(a+b)2
2
so that

a4+b4=

\left(a2\right)2
1

+

\left(b2\right)2
1

\geq

\left(a2+b2\right)2
2

\geq

\left((a+b)2\right)2
2
2

=

(a+b)4
8

.

Example 4.

We have that

1
a+b

+

1
b+c

+

1
a+c

\geq

(1+1+1)2
2(a+b+c)

=

9
2(a+b+c)

.

Notes and References

  1. Web site: Sedrakyan . Nairi . 1997 . About the applications of one useful inequality . Kvant Journal . 42–44, 97(2), Moscow .
  2. Book: Sedrakyan, Nairi . 1997 . A useful inequality . Springer International publishing . 107 . 9783319778365 .
  3. Web site: Statement of the inequality . 2018 . Brilliant Math & Science .
  4. Book: Sedrakyan, Nairi . 2018 . Algebraic inequalities . Springer International publishing . 107–109 .