Hidden Cauchy-Schwarz inequalityIf x i &gt; 0 and x i y i − z i 2 &gt; 0 for i ≤ n , then prove that n 3 ( ∑ i = 1 n x i ) ( ∑ i = 1 n y i ) − ( ∑ i = 1 n z i 2 ) ≤ ∑ i = 1 n 1 x i y i − z i 2 This is a question from the IMO 1969. At first it seems that we can assume that a i = x i y i − z i and b i = x i y i + z i . Hence RHS becomes ∑ i = 1 n 1 a i b i .Now how do we simplify the LHS so that we can use the Cauchy-Schwarz inequality?

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Hidden Cauchy-Schwarz inequalityIf       x          i        &amp;gt;  0 and       x          i            y          i        −                    z                  i                      2    &amp;gt;  0 for   i  ≤  n , then prove that            n      3              (              ∑                  i          =          1                n                    x                  i                    )      (              ∑                  i          =          1                n                    y                  i                    )      −      (              ∑                  i          =          1                n                                          z                          i                                      2            )        ≤      ∑          i      =      1        n        1                  x                  i                            y                  i                    −              z                  i                2            This is a question from the IMO 1969. At first it seems that we can assume that       a          i        =            x              i                    y              i              −      z          i       and       b          i        =            x              i                    y              i              +      z          i      . Hence RHS becomes       ∑          i      =      1        n        1                  a                  i                            b                  i                    .Now how do we simplify the LHS so that we can use the Cauchy-Schwarz inequality?

Ethen Valentine · Accepted Answer

let,       f          i        (  x  )  =      x          i            x    2    −  2      z          i        x  +      y          i      Hence,       min          x      ∈              R                  f          i        (  x  )  =                    x                  i                            y                  i                    −              z                  i                2                    x              i                  min          x      ∈              R                  ∑          i      =      1              n            f          i        (  x  )  ≥      ∑          i      =      1              n            min          x      ∈              R                  f          i        (  x  )Hence,                      ∑                  i          =          1                n                    x                  i                    ⋅              ∑                  i          =          1                          n                            y                  i                    −      (              ∑                  i          =          1                          n                            z                  i                            )        2                            ∑                  i          =          1                          n                            x                  i                      ≥      ∑          i      =      1              n                          x                  i                            y                  i                    −              z                  i                2                    x              i             and      (          ∑              i        =        1                    n                    x              i              ⋅          ∑              i        =        1                    n                    y              i              −                  (                  ∑                      i            =            1                                n                                    z                      i                          )                    2              )    ⋅      ∑          i      =      1              n            1                  x                  i                            y                  i                    −              z                  i                          2                      ≥  ≥      ∑          i      =      1              n            x          i        ⋅      ∑          i      =      1              n                          x                  i                            y                  i                    −              z                  i                          2                            x              i              ⋅      ∑          i      =      1              n            1                  x                  i                            y                  i                    −              z                  i                          2                      ≥      n          3        .

Hidden Cauchy-Schwarz inequality If x <mrow class="MJX-TeXAtom-ORD"> i

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