Olympiad Inequality ∑ c y c x 4 8 x 3 + 5 y 3 ⩾ x + y + z 13 x , y , z &gt; 0 , prove x 4 8 x 3 + 5 y 3 + y 4 8 y 3 + 5 z 3 + z 4 8 z 3 + 5 x 3 ⩾ x + y + z 13 Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".

Question

Olympiad Inequality        ∑          c      y      c                  x      4              8              x        3            +      5              y        3              ⩾            x      +      y      +      z        13    x  ,  y  ,  z  &amp;gt;  0 , prove            x      4              8              x        3            +      5              y        3              +            y      4              8              y        3            +      5              z        3              +            z      4              8              z        3            +      5              x        3              ⩾            x      +      y      +      z        13  Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don&#039;t believe that any students can solve this problem in 3 hour time frame.Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as &quot;coffin problems&quot; in my country. The official solution is very elementary and elegant.Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with &quot;pencil and papers.&quot; I think the approach by Peter Scholze in here may help.Update 3: Michael has tried to apply Peter Scholze&#039;s method but not found the solution yet.Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the &quot;King of Inequality&quot;.

Carl Norris · Accepted Answer

A big problem we get around (x,y,z)=(0.822,1.265,1.855).The Buffalo Way helps:Let x=min{x,y,z}, y=x+u,z=x+v and x=tuv−−√.Hence,       13    5        ∏          c      y      c        (  8      x    3    +  5      y    3    )      (          ∑              c        y        c                            x        4                    8                  x          3                +        5                  y          3                      −                  x        +        y        +        z            13        )    =  =  156  (      u    2    −  u  v  +      v    2    )      x    8    +  6  (  65      u    3    +  189      u    2    v  −  176  u      v    2    +  65      v    3    )      x    7    +  +  2  (  377      u    4    +  1206      u    3    v  +  585      u    2        v    2    −  1349  u      v    3    +  377      v    4    )      x    6    +  +  3  (  247      u    5    +  999      u    4    v  +  1168      u    3        v    2    −  472      u    2        v    3    −  726  u      v    4    +  247  )      x    5    +  +  3  (  117      u    6    +  696      u    5    v  +  1479      u    4        v    2    +  182      u    3        v    3    −  686      u    2        v    4    −  163  u      v    5    +  117      v    6    )      x    4    +  +  (  65      u    7    +  768      u    6    v  +  2808      u    5        v    2    +  2079      u    4        v    3    −  1286      u    3        v    4    −  585      u    2        v    5    +  181  u      v    6    +  65      v    7    )      x    3    +  +  3  u  v  (  40      u    6    +  296      u    5    v  +  472      u    4        v    2    −  225      u    2        v    4    +  55  u      v    5    +  25      v    6    )      x    2    +  +      u    2        v    2    (  120      u    5    +  376      u    4    v  +  240      u    3        v    2    −  240      u    2        v    3    −  25  u      v    4    +  75      v    5    )  x  +  +  5      u    3        v    3    (  8      u    4    +  8      u    3    v  −  8  u      v    3    +  5      v    4    )  ≥  ≥      u    5        v    5    (  156      t    8    +  531      t    7    +  2      t    6    −  632      t    5    −  152      t    4    +  867      t    3    +  834      t    2    +  299  t  +  40  )  ≥  0Done!For example, we&#039;ll prove that  6  (  65      u    3    +  189      u    2    v  −  176  u      v    2    +  65      v    3    )  ≥  531            u      3              v      3        ,which gives a coefficient 531 before       t    7   in the polynomial   156      t    8    +  531      t    7    +  2      t    6    −  632      t    5    −  152      t    4    +  867      t    3    +  834      t    2    +  299  t  +  40..Indeed, let   u  =      k    2    v, where k&amp;gt;0.Thus, we need to prove that:  130      k    6    +  378      k    4    −  177      k    3    −  352      k    2    +  130  ≥  0and by AM-GM we obtain:  130      k    6    +  378      k    4    −  177      k    3    −  352      k    2    +  130  =  =  130            (              k        3            +              10        13            k      −      1      )        2    +      k    13    (  2314      k    3    +  1079      k    2    −  5576  k  +  2600  )  ≥  ≥      k    13        (    8    ⋅          1157      4              k      3        +    5    ⋅          1079      5              k      2        +    21    ⋅          2600      21        −    5576    k    )    ≥  ≥            k      2        13        (    34                                        (                          1157              4                        )                    8                                      (                          1079              5                        )                    5                                      (                          2600              21                        )                                21                              34        −    5576    )    &amp;gt;  0.We&#039;ll prove that  2  (  377      u    4    +  1206      u    3    v  +  585      u    2        v    2    −  1349  u      v    3    +  377      v    4    )  ≥  2      u    2        v    2    ,for which it&#039;s enough to prove that:  377      t    4    +  1206      t    3    +  584      t    2    −  1349  t  +  377  ≥  0or      t    4    +      1206    377        t    3    +      584    377        t    2    −      1349    377    t  +  1  ≥  0or            (              t        2            +              603        377            t      −              28        29            )        2    +            131015              t        2            −      69589      t      +      9633        142129    ≥  0  ,which is true because      69589    2    −  4  ⋅  131015  ⋅  9633  &amp;lt;  0.

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