Prove that 1 a b + a + 2 + 1 b c + b + 2 + 1 a c + c + 2 ≤ 3 2 Let a b c = 1 and a , b , c &gt; 0. Prove that 1 a b + a + 2 + 1 b c + b + 2 + 1 a c + c + 2 ≤ 3 2 .I guess that 1 a b + a + 2 + 1 b c + b + 2 + 1 a c + c + 2 = 3 2 if and only if a = b = c = 1. I have spent two nights on this but it's just a mess and I still don't know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.

Question

Prove that       1          a      b      +      a      +      2        +      1          b      c      +      b      +      2        +      1          a      c      +      c      +      2        ≤      3    2  Let   a  b  c  =  1 and   a  ,  b  ,  c  &amp;gt;  0. Prove that      1          a      b      +      a      +      2        +      1          b      c      +      b      +      2        +      1          a      c      +      c      +      2        ≤      3    2    .I guess that      1          a      b      +      a      +      2        +      1          b      c      +      b      +      2        +      1          a      c      +      c      +      2        =      3    2  if and only if   a  =  b  =  c  =  1. I have spent two nights on this but it&#039;s just a mess and I still don&#039;t know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.

Kason Gonzales · Accepted Answer

Hint:  Let      1          a      b      +      a      +      2        +      1          b      c      +      b      +      2        +      1          a      c      +      c      +      2        =  AApplying C-S:      (          1              a        b        +        a        +        2              +          1              b        c        +        b        +        2              +          1              a        c        +        c        +        2              )    (  3  )  ≥  (  A      )    2  Hence  A  ≤            (              1                  a          b          +          a          +          2                    +              1                  b          c          +          b          +          2                    +              1                  a          c          +          c          +          2                    )        (    3    )  Now consider by Titu&#039;s lemma we have      1          a      b      +      1        +      1          a      +      1        ≥      4          a      b      +      a      +      2      Hence our obtained inequality now has  A  ≤            (              1                  a          b          +          a          +          2                    +              1                  b          c          +          b          +          2                    +              1                  a          c          +          c          +          2                    )        (    3    )    ≤            3      4              ∑              c        y        c                    (              1                  a          b          +          1                    +              1                  a          +          1                    )      But      ∑          c      y      c            (          1              a        b        +        1              +          1              a        +        1              )    =      ∑          c      y      c            (          c              c        +        1              +          1              a        +        1              )    =  3Because   a  b  c  =  1 Hence  A  ≤            3      4              ∑              c        y        c                    c              c        +        1              +          1              a        +        1              =            9      4        =      3    2  Hope it helped now.

Kamila Frye · Accepted Answer

Let   a  =      y    x   and   b  =      z    y  , where   x,   y and   z are positives.Thus,   c  =      x    z   and by C-S we obtain:      ∑          c      y      c            1          a      b      +      a      +      2        =      ∑          c      y      c                  1                        z          x                +                  y          x                +        2              =      ∑          c      y      c                  x              2        x        +        y        +        z              ≤            ∑              c        y        c                    1      2              ∑              c        y        c                    x              2        x        +        y        +        z              =      3          ∑              c        y        c              x    ⋅          1              x        +        z        +        x        +        y              ≤      3          ∑              c        y        c                    x              (        1        +        1                  )          2                            (                        1          2                          x          +          z                    +                        1          2                          x          +          y                    )        =            3      4              ∑              c        y        c                    (              x                  x          +          z                    +              x                  x          +          y                    )        =            3      4              ∑              c        y        c                    (              y                  y          +          x                    +              x                  x          +          y                    )        =      3    2    .

Prove that 1/(sqrt(ab+a+2))+ 1/(sqrt(bc+b+2))+ 1/(sqrt(ac+c+2)) <= 3/2

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