A four variable inequalityLet a , b , c , d be non-negative real numbers satisfying a + b + c + d = 3. Show a 1 + 2 b 3 + b 1 + 2 c 3 + c 1 + 2 d 3 + d 1 + 2 a 3 ⩾ a 2 + b 2 + c 2 + d 2 3 .I got this inequality from an IMO preparation club. I no longer do contest math but still I find this interesting. There is a weird equality case : ( 2 , 1 , 0 , 0 ) works (whereas ( 2 , 0 , 1 , 0 ) does not).So far I've tried regrouping terms to disminish the number of variables or the number of non-zero variables. I've shown that the inequality is implied by the three-variable version a 1 + 2 b 3 + b 1 + 2 c 3 + c 1 + 2 a 3 ⩾ a 2 + b 2 + c 2 3 if a + b + c = 3. The three variable version itself is implied by the two variable version, but the two variable version is false.

Question

A four variable inequalityLet   a  ,  b  ,  c  ,  d be non-negative real numbers satisfying   a  +  b  +  c  +  d  =  3. Show      a          1      +      2              b        3              +      b          1      +      2              c        3              +      c          1      +      2              d        3              +      d          1      +      2              a        3              ⩾                    a        2            +              b        2            +              c        2            +              d        2              3    .I got this inequality from an IMO preparation club. I no longer do contest math but still I find this interesting. There is a weird equality case :   (  2  ,  1  ,  0  ,  0  ) works (whereas   (  2  ,  0  ,  1  ,  0  ) does not).So far I&#039;ve tried regrouping terms to disminish the number of variables or the number of non-zero variables. I&#039;ve shown that the inequality is implied by the three-variable version      a          1      +      2              b        3              +      b          1      +      2              c        3              +      c          1      +      2              a        3              ⩾                    a        2            +              b        2            +              c        2              3  if   a  +  b  +  c  =  3. The three variable version itself is implied by the two variable version, but the two variable version is false.

Xavier Jennings · Accepted Answer

By AM-GM      ∑          c      y      c            a          1      +      2              b        3              =  3  +      ∑          c      y      c            (          a              1        +        2                  b          3                      −    a    )    =  3  −      ∑          c      y      c                  2      a              b        3                    1      +      2              b        3              ≥  ≥  3  −      ∑          c      y      c                  2      a              b        3                    3              b        2              =            (      a      +      b      +      c      +      d              )        2            −      2              ∑                  c          y          c                    a      b        3    =  =                    a        2            +              b        2            +              c        2            +              d        2            +      a      c      +      b      d        3    ≥                    a        2            +              b        2            +              c        2            +              d        2              3    .

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