Showing 1 2 3 2 − p ( x + y + z ) p − 1 ≤ x p y + z + y p x + z + z p y + x with p &gt; 1, and x , y , z positiveBy Jensen I got taking p 1 = p 2 = p 3 = 1 3 , (say x = x 1 , y = x 2 , z = x 3 ) ( ∑ p k x k ) p ≤ ∑ p k x p which means; 3 1 − p ( x + y + z ) p ≤ x p + y p + z p and if I assume wlog x ≥ y ≥ z then 1 y + z ≥ 1 x + z ≥ 1 x + y so these 2 sequences are similarly ordered, and by rearrangement inequality the rest follows ?I should use power means, but any other solution is also appreciated.

Question

Showing       1    2        3          2      −      p        (  x  +  y  +  z      )          p      −      1        ≤            x      p              y      +      z        +            y      p              x      +      z        +            z      p              y      +      x       with   p  &amp;gt;  1, and   x  ,  y  ,  z positiveBy Jensen I got taking       p    1    =      p    2    =      p    3    =      1    3  , (say   x  =      x    1    ,  y  =      x    2    ,  z  =      x    3  )            (      ∑              p        k                    x        k            )        p    ≤  ∑      p    k        x    p   which means;      3          1      −      p        (  x  +  y  +  z      )    p    ≤      x    p    +      y    p    +      z    p  and if I assume wlog   x  ≥  y  ≥  z then       1          y      +      z        ≥      1          x      +      z        ≥      1          x      +      y      so these   2 sequences are similarly ordered, and by rearrangement inequality the rest follows ?I should use power means, but any other solution is also appreciated.

Cael Cox · Accepted Answer

Indeed, Jensen works.Let   x  +  y  +  z  =  3Hence, we need to prove that       ∑          c      y      c        f  (  x  )  ≥  0, where   f  (  x  )  =            x      p              3      −      x        −      1    2        f    ″    (  x  )  =                    x                  p          −          2                            (        (        p        −        2        )        (        p        −        1        )                  x          2                −        6        p        (        p        −        2        )        x        +        9        p        (        p        −        1        )        )                    (      3      −      x              )        3            If   p  =  2 so       f    ″    (  x  )  &amp;gt;  0If   p  &amp;gt;  2 so since       f    ″    (  0  )  &amp;gt;  0 so since       f    ″    (  0  )  &amp;gt;  0 and       lim          x      →              3        −                  f    ″    (  x  )  &amp;gt;  0 and             3      p              p      −      1        &amp;gt;  3, we see that       f    ″    (  x  )  &amp;gt;  0If   1  &amp;lt;  p  &amp;lt;  2 so since       f    ″    (  0  )  &amp;gt;  0 and       lim          x      →              3        −                  f    ″    (  x  )  &amp;gt;  0, we see that       f    ″    (  x  )  &amp;gt;  0Thus, your inequality follows from Jensen.Done

Ibrahim Rosales · Accepted Answer

WLOG,                    (1)                    x        +        y        +        z        =        1            then by power mean inequality;      ∑          c      y      c                  (              1        3            (      x      +      y              )                  −          1                    )              −      1        ≤  (  x  +  y      )                  1        3              (  y  +  z      )                  1        3              (  x  +  z      )                  1        3            and AM-GM  (  x  +  y      )                  1        3              (  y  +  z      )                  1        3              (  x  +  z      )                  1        3              ≤      1    3    (  x  +  y  )  +      1    3    (  y  +  z  )  +      1    3    (  x  +  z  )      =          (      1      )            2    3            ⟹              1              x        +        y              +          1              y        +        z              +          1              x        +        z              ≥          9      2      Rearrangement Inequality implies,                    x        p                    y        +        z              +                  y        p                    x        +        z              +                  z        p                    y        +        x              ≥                  (                  x          p                +                  y          p                +                  z          p                )        ⋅                  (                      1                          x              +              y                                +                      1                          y              +              z                                +                      1                          x              +              z                                )                    3      and in general one has             (      ∑              p        k                    x        k            )        p    ≤  ∑      p    k        x    p   thus,      3          p      −      1        (      x    p    +      y    p    +      z    p    )  ≥  (  x  +  y  +  z      )          p            =          (      1      )        (  x  +  y  +  z      )          p      −      1                ⟹                      x        p                    y        +        z              +                  y        p                    x        +        z              +                  z        p                    y        +        x              ≥                  (                  x          p                +                  y          p                +                  z          p                )        ⋅                  9          2                    3        ≥                            3                      1            −            p                          (        x        +        y        +        z                  )                      p            −            1                          ⋅                  9          2                    3        =          1      2              3              2        −        p              (    x    +    y    +    z          )              p        −        1

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