For a b c = 1 prove that ∑ cyc 1 a + 3 ≥ ∑ cyc a a 2 + 3 I tried TL, BW, the Vasc's Theorems and more, but without success.I proved this inequality!I proved also the hardest version: ∑ c y c 1 a + 4 ≥ ∑ c y c a a 2 + 4 Thanks all!

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For   a  b  c  =  1 prove that       ∑    cyc        1          a      +      3        ≥      ∑    cyc        a                  a        2            +      3      I tried TL, BW, the Vasc&#039;s Theorems and more, but without success.I proved this inequality!I proved also the hardest version:       ∑          c      y      c            1          a      +      4        ≥      ∑          c      y      c            a                  a        2            +      4      Thanks all!

Nolan Tyler · Accepted Answer

BW in the following version does not help.Let   a  =      x    3  ,   b  =      y    3   and   c  =      z    3  Hence, we need to prove that      ∑          c      y      c            1                  x        3            +      3      x      y      z        ≥      ∑          c      y      c                  x      3                      x        6            +      3              x        2                    y        2                    z        2            or      ∑          c      y      c            1                  x        3            +      3      x      y      z        ≥      ∑          c      y      c            x                  x        4            +      3              y        2                    z        2              .Now, we can assume that   x  =  min  {  x  ,  y  ,  z  },   y  =  x  +  u and   z  =  x  +  vand these substitutions give inequality, which I don&#039;t know to prove.But we can use another BW!Let   a  =      y    x  ,   b  =      z    y   and   c  =      x    z  , where    x,   y and   z are positives.Hence, we need to prove that      ∑          c      y      c            x          3      x      +      y        ≥      ∑          c      y      c                  x      y              3              x        2            +              y        2            or      ∑          c      y      c                          x        3            −              x        2            y              (      3      x      +      y      )      (      3              x        2            +              y        2            )        ≥  0.Now, let   x  =  min  {  x  ,  y  ,  z  },   y  =  x  +  u and   z  =  x  +  vHence, we need to prove that  128  (      u    2    −  u  v  +      v    2    )      x    7    +  16  (  16      u    3    +  23      u    2    v  −  15  u      v    2    +  16      v    3    )      x    6    +  +  32  (  8      u    4    +  27      u    3    v  +  12      u    2        v    2    −  11  u      v    3    +  8      v    4    )      x    5    +  +  4  (  32      u    5    +  193      u    4    v  +  266      u    3        v    2    −  42      u    2        v    3    −  33  u      v    4    +  32      v    5    )      x    4    +  +  2  (  8      u    6    +  178      u    5    v  +  435      u    4        v    2    +  152      u    3        v    3    −  99      u    2        v    4    +  30  u      v    5    +  8      v    6    )      x    3    +  +  u  v  (  45      u    5    +  375      u    4    v  +  291      u    3        v    2    −  83      u    2        v    3    +  57  u      v    4    +  3      v    5    )      x    2    +  +  2      u    2        v    2    (  24      u    4    +  66      u    3    v  −  18      u    2        v    2    +  13  u      v    3    +  3      v    4    )  x  +  +      u    3        v    3    (  18      u    3    −  6      u    2    v  +  3  u      v    2    +      v    3    )  ≥  0  ,which is obvious.Done!

For abc=1 prove that sum lim_text{cyc}(1)/(a+3)>=sum lim_text{cyc}(a)/(a^2+3) Let a, b and c be positive numbers such that abc=1. Prove that: 1/(a+3)+1/(b+3)+1/(c+3)>=a/(a^2+3)+b/(b^2+3)+c/(c^2+3)

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