Sedrakyan's inequality

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The following inequality is known as Sedrakyan's inequality, Bergström's inequality, Engel's form or Titu's lemma, 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 (mathematician) 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 (Nairi Sedrakyan (1997), Arthur Engel (mathematician) (1998), Titu Andreescu (2003))[]

For any reals ${\displaystyle a_{1},a_{2},a_{3},\ldots ,a_{n}}$ and positive reals ${\displaystyle b_{1},b_{2},b_{3},\ldots ,b_{n},}$ we have ${\displaystyle {\frac {a_{1}^{2}}{b_{1}}}+{\frac {a_{2}^{2}}{b_{2}}}+\cdots +{\frac {a_{n}^{2}}{b_{n}}}\geq {\frac {\left(a_{1}+a_{2}+\cdots +a_{n}\right)^{2}}{b_{1}+b_{2}+\cdots +b_{n}}}.}$

Direct applications[]

Example 1. Nesbitt's inequality.

For positive real numbers ${\displaystyle a,b,c:}$ ${\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}$

Example 2. International Mathematical Olympiad (IMO) 1995.

For positive real numbers ${\displaystyle a,b,c}$, where ${\displaystyle abc=1}$ we have that ${\displaystyle {\frac {1}{a^{3}(b+c)}}+{\frac {1}{b^{3}(a+c)}}+{\frac {1}{c^{3}(a+b)}}\geq {\frac {3}{2}}.}$

Example 3.

For positive real numbers ${\displaystyle a,b}$ we have that ${\displaystyle 8(a^{4}+b^{4})\geq (a+b)^{4}.}$

Example 4.

For positive real numbers ${\displaystyle a,b,c}$ we have that ${\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {9}{2(a+b+c)}}.}$

Proofs[]

Example 1.

Proof: Use ${\displaystyle n=3,}$ ${\displaystyle \left(a_{1},a_{2},a_{3}\right):=(a,b,c),}$ and ${\displaystyle \left(b_{1},b_{2},b_{3}\right):=(a(b+c),b(c+a),c(a+b))}$ to conclude:

$${\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {(a+b+c)^{2}}{a(b+c)+b(c+a)+c(a+b)}}={\frac {a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}{2(ab+bc+ca)}}={\frac {a^{2}+b^{2}+c^{2}}{2(ab+bc+ca)}}+1\geq {\frac {1}{2}}(1)+1={\frac {3}{2}}.\blacksquare }$$

Example 2.

We have that ${\displaystyle {\frac {{\Big (}{\frac {1}{a}}{\Big )}^{2}}{a(b+c)}}+{\frac {{\Big (}{\frac {1}{b}}{\Big )}^{2}}{b(a+c)}}+{\frac {{\Big (}{\frac {1}{c}}{\Big )}^{2}}{c(a+b)}}\geq {\frac {{\Big (}{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}{\Big )}^{2}}{2(ab+bc+ac)}}={\frac {ab+bc+ac}{2}}\geq {\frac {3{\sqrt[{3}]{a^{2}b^{2}c^{2}}}}{2}}={\frac {3}{2}}.}$

Example 3.

We have ${\displaystyle {\frac {a^{2}}{1}}+{\frac {b^{2}}{1}}\geq {\frac {(a+b)^{2}}{2}}}$ so that ${\displaystyle a^{4}+b^{4}={\frac {\left(a^{2}\right)^{2}}{1}}+{\frac {\left(b^{2}\right)^{2}}{1}}\geq {\frac {\left(a^{2}+b^{2}\right)^{2}}{2}}\geq {\frac {\left({\frac {(a+b)^{2}}{2}}\right)^{2}}{2}}={\frac {(a+b)^{4}}{8}}.}$

Example 4.

We have that ${\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {(1+1+1)^{2}}{2(a+b+c)}}={\frac {9}{2(a+b+c)}}.}$

References[]