Proving an inequality via the Cauchy-Schwarz InequalityI apolgize for contributing yet another question asking about an application of CS. Here it is:Suppose p 1 , … , p n and a 1 , . . . , a n are real numbers such that p i ≥ 0, a i ≥ 0 for all i, and p 1 + ⋯ + p n = 1. Then ( p 1 a 1 + ⋯ + p n a n ) ( p 1 a 1 + ⋯ + p n a n ) ≥ 1The author of my textbook gives the following proof: Apply Cauchy's inequality to the sequences p 1 a 1 … p n a n and p 1 a 1 … p n a n . (Thats it)In trying to fill in the blanks I obtained the following p 1 a 1 + ⋯ + p n a n ≤ p 1 + ⋯ + p n a 1 + . . . + a n = a 1 + . . . + a n and p 1 a 1 + ⋯ + p n a n ≤ p 1 + ⋯ + p n 1 a 1 + . . . + 1 a n = 1 a 1 + . . . + 1 a n I'm not entirely sure where to go from here. Perhaps I have misunderstood what he meant by "apply cauchy's inequality to the sequences...". Another idea I had was to note that ( p 1 a 1 + ⋯ + p n a n ) ≤ M a ( p 1 + ⋯ + p n )where M a is the largest a i . And, that ( p 1 a 1 + ⋯ + p n a n ) ≤ 1 m a ( p 1 + ⋯ + p n )where m a is the smallest a i . Therefore, since M a m a ≥ 1 the inequality follows. I am not very confident in the correctness of this method though and would like to understand how to prove the inequality via CS as my book suggests.

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Proving an inequality via the Cauchy-Schwarz InequalityI apolgize for contributing yet another question asking about an application of CS. Here it is:Suppose       p    1    ,  …  ,      p    n   and       a    1    ,  .  .  .  ,      a    n   are real numbers such that       p    i    ≥  0,       a    i    ≥  0 for all   i, and       p    1    +  ⋯  +      p    n    =  1. Then  (      p    1        a    1    +  ⋯  +      p    n        a    n    )      (                  p        1                    a        1              +    ⋯    +                  p        n                    a        n              )    ≥  1The author of my textbook gives the following proof: Apply Cauchy&#039;s inequality to the sequences             p      1              a      1        …            p      n                  a      n       and                     p        1                    a        1              …                    p        n                    a        n            . (Thats it)In trying to fill in the blanks I obtained the following            p      1              a      1        +  ⋯  +            p      n              a      n        ≤            p      1        +    ⋯    +          p      n                  a      1        +    .    .    .    +          a      n        =            a      1        +    .    .    .    +          a      n      and                    p        1                    a        1              +  ⋯  +                    p        n                    a        n              ≤            p      1        +    ⋯    +          p      n                  1              a        1              +    .    .    .    +          1              a        n              =            1              a        1              +    .    .    .    +          1              a        n            I&#039;m not entirely sure where to go from here. Perhaps I have misunderstood what he meant by &quot;apply cauchy&#039;s inequality to the sequences...&quot;. Another idea I had was to note that  (      p    1        a    1    +  ⋯  +      p    n        a    n    )  ≤      M    a    (      p    1    +  ⋯  +      p    n    )where       M    a   is the largest       a    i  . And, that      (                  p        1                    a        1              +    ⋯    +                  p        n                    a        n              )    ≤      1          m      a        (      p    1    +  ⋯  +      p    n    )where       m    a   is the smallest       a    i  . Therefore, since             M      a              m      a        ≥  1 the inequality follows. I am not very confident in the correctness of this method though and would like to understand how to prove the inequality via CS as my book suggests.

broeifl · Accepted Answer

C-S is the following.  (      a    1    2    +      a    2    2    +  .  .  .  +      a    n    2    )  (      b    1    2    +      b    2    2    +  .  .  .  +      b    n    2    )  ≥  (      a    1        b    1    +      a    2        b    2    +  .  .  .  +      a    n        b    n        )    2    .Hence, for positives       a    i   and your       p    i   we obtain:  (      p    1        a    1    +  ⋯  +      p    n        a    n    )      (                  p        1                    a        1              +    ⋯    +                  p        n                    a        n              )    =      ∑          i      =      1        n              (                        p          i                          a          i                    )        2        ∑          i      =      1        n              (                                    p            i                                a            i                              )        2    ≥  ≥            (              ∑                  i          =          1                n            (                        p          i                          a          i                                                  p            i                                a            i                              )        2    =            (              ∑                  i          =          1                n                    p        i            )        2    =  1

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