Prove the inequalitiyIf a , b , c are positive reals such that: a b + c + 1 + b a + c + 1 + c b + a + 1 ≤ 1 then: 1 b + c + 1 + 1 a + c + 1 + 1 b + a + 1 ≥ 1I tried using inequality of arithmetic and harmonic means, to no avail.Any help is appreciated.

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Prove the inequalitiyIf   a  ,  b  ,  c are positive reals such that:       a          b      +      c      +      1        +      b          a      +      c      +      1        +      c          b      +      a      +      1        ≤  1 then:      1          b      +      c      +      1        +      1          a      +      c      +      1        +      1          b      +      a      +      1        ≥  1I tried using inequality of arithmetic and harmonic means, to no avail.Any help is appreciated.

cefflid6y · Accepted Answer

We&#039;ll write the condition in the following form:  a  +  b  +  c  +  1  ≥      a    3    +      b    3    +      c    3    +  a  b  c  +      ∑          c      y      c        (      a    2    −  a  b  )and since       ∑          c      y      c        (      a    2    −  a  b  )  ≥  0 we obtain  a  +  b  +  c  +  1  ≥      a    3    +      b    3    +      c    3    +  a  b  cIn another hand, we need to prove that  2  (  a  +  b  +  c  +  1  )  ≥  (  a  +  b  )  (  a  +  c  )  (  b  +  c  )Thus, it remains to prove that  2  (      a    3    +      b    3    +      c    3    +  a  b  c  )  ≥  (  a  +  b  )  (  a  +  c  )  (  b  +  c  )which is       ∑          c      y      c        (  a  +  b  )  (  a  −  b      )    2    ≥  0. Done!

Rapsinincke · Accepted Answer

.WLOG,   a  &amp;lt;  b  &amp;lt;  c, then       1          b      +      c        &amp;lt;      1          a      +      c        &amp;lt;      1          a      +      b      Note that      a          b      +      c      +      1        +      b          a      +      c      +      1        +      c          a      +      b      +      1        =  (  a  +  b  +  c  +  1  )            (            1          b      +      c      +      1        +      1          a      +      c      +      1        +      1          a      +      b      +      1                  )        −  3Hence,  (  a  +  b  +  c  +  1  )            (            1          b      +      c      +      1        +      1          a      +      c      +      1        +      1          a      +      b      +      1                  )        ≤  4Further, by the harmonic mean-arithmetic mean inequality,            (            1          b      +      c      +      1        +      1          a      +      c      +      1        +      1          a      +      b      +      1                  )        ≥            9              2        (        a        +        b        +        c        )        +        3            Now, let   x  =            (            1          b      +      c      +      1        +      1          a      +      c      +      1        +      1          a      +      b      +      1                  )       and   y  =  a  +  b  +  c. Then, we have:   2  y  x  +  3  x  ≥  9 and   4  ≥  y  x  +  x. Multiplying by   2 and changing the sign,   −  2  y  x  −  2  x  ≥  8. Adding ,we get   x  ≥  1. Hence the result.

Prove the inequalitiy If a , b , c are positive reals such that: a

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