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]
For any real numbers
and positive reals
we have
(Nairi Sedrakyan (1997), Arthur Engel (1998), Titu Andreescu (2003))
Example 1. Nesbitt's inequality.
For positive real numbers
![{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/ec8cddcaf162a1509f22053e879df555a0c28ab1)
Example 2. International Mathematical Olympiad (IMO) 1995.
For positive real numbers
, where
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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/0ee2aa3452f6ba9dc4bfcf474d9096b4908f8f40)
Example 3.
For positive real numbers
we have that ![{\displaystyle 8(a^{4}+b^{4})\geq (a+b)^{4}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/c44bcbd1c34425aaf166d76cb49bfe480914de11)
Example 4.
For positive real numbers
we have that ![{\displaystyle {\frac {1}{a+b}}+{\frac {1}{b+c}}+{\frac {1}{a+c}}\geq {\frac {9}{2(a+b+c)}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/864a24dc44284140b64bedf8b8b46b2ef3696e87)
Proofs
Example 1.
Proof: Use
and
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 }](//wikimedia.org/api/rest_v1/media/math/render/svg/d1dbaa8468df4268b610e24a9e67ab98e807bde5)
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}{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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/6bbe2468a62831c6241aed12e751dd3694543884)
Example 3.
We have
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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/6e637f569fc378413dc1ec800dea4a2c97282b0d)
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)}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/7d85084bd849fbab9509f02871da32cb12d80a08)