Nice inequality with a,b,c,dIf a , b , c , d &gt; 0 with a + b + c + d = 1, prove that b c d a + 2 + a c d b + 2 + a b d c + 2 + a b c d + 2 &lt; 1 13

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Nice inequality with a,b,c,dIf   a  ,  b  ,  c  ,  d  &amp;gt;  0 with   a  +  b  +  c  +  d  =  1, prove that                     b        c        d                    a        +        2              +                  a        c        d                    b        +        2              +                  a        b        d                    c        +        2              +                  a        b        c                    d        +        2              &amp;lt;          1      13

ri1men4dp · Accepted Answer

By AM-GM we obtain:      ∑          c      y      c                  b      c      d              a      +      2        =      1    2        ∑          c      y      c        b  c  d  +      ∑          c      y      c            (                  b        c        d                    a        +        2              −                  b        c        d            2        )    =  =      1    2        ∑          c      y      c        b  c  d  −      ∑          c      y      c                  a      b      c      d              2      (      a      +      2      )        &amp;lt;      1    2        ∑          c      y      c        b  c  d  =  =            a      b      c      +      a      b      d      +      a      c      d      +      b      c      d        2    =            a      b      (      c      +      d      )      +      c      d      (      a      +      b      )        2    ≤  ≤                              (                                    a              +              b                        2                    )                2            (      c      +      d      )      +                        (                                    c              +              d                        2                    )                2            (      a      +      b      )        2    =            (      a      +      b      )      (      c      +      d      )      (      a      +      b      +      c      +      d      )        8    =  =            (      a      +      b      )      (      c      +      d      )        8    ≤                    (                              a            +            b            +            c            +            d                    2                )            2        8    =      1    32    &amp;lt;      1    13

John Landry · Accepted Answer

Not so clear and &quot;good&quot; bound solution, but get necessary:  (  1  ) Use Cauchy inequality   b  c  d  ⩽  (                    b        +        c        +        d            3            )    3  ;   a  c  d  ⩽  (                    a        +        c        +        d            3            )    3  ;   a  b  d  ⩽  (                    a        +        b        +        d            3            )    3  ;   a  b  c  ⩽  (                    a        +        b        +        c            3            )    3  as   a  +  b  +  c  +  d  =  1, we get   b  c  d  ⩽  (                    1        −        a            3            )    3  ;   a  c  d  ⩽  (                    1        −        b            3            )    3  ;   a  b  d  ⩽  (                    1        −        c            3            )    3  ;   a  b  c  ⩽  (                    1        −        d            3            )    3  So      ∑          c      y      c                  b      c      d              a      +      2        ⩽                    (        1        −        a                  )          3                            27        (        a        +        2        )              +                    (        1        −        b                  )          3                            27        (        b        +        2        )              +                    (        1        −        c                  )          3                            27        (        c        +        2        )              +                    (        1        −        d                  )          3                            27        (        d        +        2        )            Then, as   0  &amp;lt;  a  ,  b  ,  c  ,  d  &amp;lt;  1 we can take value at zero in the right side inequality addends for further inequality estimation, because numerator increases and denominator decreases as   a  ,  b  ,  c  ,  d approach to   0Finally, we have:      ∑          c      y      c                  b      c      d              a      +      2        ⩽                    (        1        −        a                  )          3                            27        (        a        +        2        )              +                    (        1        −        b                  )          3                            27        (        b        +        2        )              +                    (        1        −        c                  )          3                            27        (        c        +        2        )              +                    (        1        −        d                  )          3                            27        (        d        +        2        )              &amp;lt;      1    54    +      1    54    +      1    54    +      1    54    =      1    13.5    &amp;lt;      1    13

Nice inequality with a,b,c,d If a,b,c,d>0 with a+b+c+d=1, prove that (bcd)/(a+2)+(acd)/(b+2)+(abd)/(c+2)+(abc)/(d+2)<(1)/(13)

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