Inequality with a+b+c=1Let a , b , c be positive reals such that a + b + c = 1. Prove that b c a + 5 + c a b + 5 + a b c + 5 ≤ 1 4 .Progress: This is equivalent to show that 1 a 2 + 5 a + 1 b 2 + 5 b + 1 c 2 + 5 c ≤ 1 4 a b c .I'm not sure how to proceed further. Also equality doesn't occur when a = b = c, which makes me more confused.Edit: The original inequality is as follows: Let a 1 , a 2 , ⋯ , a n be n &gt; 2 positive reals such that a 1 + a 2 + ⋯ + a n = 1. Prove that, ∑ k = 1 n ∏ j ≠ k a j a k + n + 2 ≤ 1 ( n − 1 ) 2 .

Question

Inequality with a+b+c=1Let   a  ,  b  ,  c be positive reals such that   a  +  b  +  c  =  1. Prove that                    b        c                    a        +        5              +                    c        a                    b        +        5              +                    a        b                    c        +        5              ≤            1      4        .Progress: This is equivalent to show that            1                        a          2                +        5        a              +            1                        b          2                +        5        b              +            1                        c          2                +        5        c              ≤            1              4        a        b        c              .I&#039;m not sure how to proceed further. Also equality doesn&#039;t occur when   a  =  b  =  c, which makes me more confused.Edit: The original inequality is as follows: Let       a    1    ,      a    2    ,  ⋯      ,      a    n   be   n  &amp;gt;  2 positive reals such that       a    1    +      a    2    +  ⋯      +      a    n    =  1. Prove that,      ∑          k      =      1        n                                ∏                      j            ≠            k                                    a          j                                      a          k                +        n        +        2              ≤            1              (        n        −        1                  )          2                      .

Miya Swanson · Accepted Answer

Notice that the conditions imply that   0  ≤  a  ,  b  ,  c  ≤  1; in particular, we have                    b        c                    a        +        5              +                    c        a                    b        +        5              +                    a        b                    c        +        5              ≤                    b        c            5        +                    c        a            5        +                    a        b            5        ≤            b      +      c      +      a        5    &amp;lt;            1      4        .I don&#039;t think this works for the general case though, because for   n  ≥  4, we have   (  n  −  1      )    2    ≥  n  +  2Now for the general case, by AM-GM inequality      ∏          j      ≠      k            a    j    ≤            (                                    ∑                          j              ≠              k                                            a            j                                    n          −          1                    )              n      −      1        =            (                        1          −                      a            k                                    n          −          1                    )              n      −      1        ≤      1          (      n      −      1              )                  n          −          1                      ,and thus      ∑          k      =      1        n                                ∏                      j            ≠            k                                    a          j                                      a          k                +        n        +        2              ≤      n          (      n      +      2      )      (      n      −      1              )                  n          −          1                      ≤      1          (      n      −      1              )                  n          −          1                      ≤            1              (        n        −        1                  )          2                    since   n  −  1  ≥  2

ct1a2n4k · Accepted Answer

We&#039;ll prove a stronger inequality:       ∑          c      y      c                  b      c              5      +      a        ≤      1    16  Indeed, we need to prove that      ∑          c      y      c                  b      c              6      a      +      5      b      +      5      c        ≤            a      +      b      +      c        16  or      ∑          c      y      c            (          a      16        −                  b        c                    6        a        +        5        b        +        5        c              )    ≥  0or      ∑          c      y      c                  6              a        2            +      5      a      b      +      5      a      c      −      16      b      c              6      a      +      5      b      +      5      c        ≥  0or      ∑          c      y      c                  (      a      −      b      )      (      3      a      +      8      c      )      −      (      c      −      a      )      (      3      a      +      8      b      )              6      a      +      5      b      +      5      c        ≥  0or      ∑          c      y      c        (  a  −  b  )      (                  3        a        +        8        c                    6        a        +        5        b        +        5        c              −                  3        b        +        8        c                    6        b        +        5        a        +        5        c              )    ≥  0or      ∑          c      y      c                  (      a      −      b              )        2            (      15      a      +      15      b      +      7      c      )              (      6      a      +      5      b      +      5      c      )      (      6      b      +      5      a      +      5      c      )        ≥  0.Done!

Let a,b,c be positive reals such that a+b+c=1. Prove that (bc)/(a+5)+(ca)/(b+5)+(ab)/(c+5)<=1/4. Progress: This is equivalent to show that (1)/(a^2+5a)+(1)/(b^2+5b)+(1)/(c^2+5c)<=(1)/(4abc). I'm not sure how to proceed further. Also equality doesn't occur when a=b=c, which makes me more confused.

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