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, 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. 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.
Statement of the inequality For any real numbers 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 Similarly to the Cauchy–Schwarz inequality, one can generalize Sedrakyan's inequality to random variables. 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 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)}}.}
Proofs 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)}}.}
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