Let f ( x ) be a differential and an invertible function, such that f ″ ( x ) &gt; 0 and f ′ ( x ) &gt; 0 .Prove f − 1 ( x 1 + x 2 + x 3 3 ) &gt; f − 1 ( x 1 ) + f − 1 ( x 2 ) + f − 1 ( x 3 ) 3

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Let       f    (    x    )   be a differential and an invertible function, such that             f      ″        (    x    )    &amp;gt;    0   and             f      ′        (    x    )    &amp;gt;    0  .Prove            f              −        1                    (                                    x            1                    +                      x            2                    +                      x            3                          3            )        &amp;gt;                            f                      −            1                          (                  x          1                )        +                  f                      −            1                          (                  x          2                )        +                  f                      −            1                          (                  x          3                )            3

wellnesshaus4n4 · Accepted Answer

I guess   f is defined on       R   and       x    1    ,      x    2    ,      x    3   are not equals. Since       f    ″    (  x  )  &amp;gt;  0,   f is stricly convex, hence  f      (                            f                      −            1                          (                  x          1                )            3        +                            f                      −            1                          (                  x          3                )            3        +                            f                      −            1                          (                  x          3                )            3        )    &amp;lt;                    x        1            +              x        2            +              x        3              3    ,and since   f is strictly increasing, so is       f          −      1      . It gives                    f                  −          1                    (              x        1            )        3    +                    f                  −          1                    (              x        3            )        3    +                    f                  −          1                    (              x        3            )        3    &amp;lt;      f          −      1            (                            x          1                +                  x          2                +                  x          3                    3        )    .

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