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 mathematical proof technique and it has very useful new applications. In the book Algebraic Inequalities (Sedrakyan) are provided several generalizations of this inequality.
 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))
 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
  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:   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)}}.}