Show that if 1 a + 1 b + 1 c = a + b + c, then 1 3 + a + 1 3 + c + 1 3 + c ≤ 3 4 The corresponding problem replacing the 3s with 2 is shown here:How to prove 1 2 + a + 1 2 + b + 1 2 + c ≤ 1?The proof is not staggeringly difficult: The idea is to show (the tough part) that if 1 2 + a + 1 2 + c + 1 2 + c = 1 then 1 a + 1 b + 1 c ≥ a + b + c. The result then easily follows.A pair of observations, both under the constraint of 1 a + 1 b + 1 c = a + b + c:If k ≥ 2 then 1 k + a + 1 k + c + 1 k + c ≤ 3 k + 1 . This is proven for k = 2 but I don't see how for k &gt; 2If 0 &lt; k &lt; 2 then there are positive ( a , b , c ) satisfying the constraint such that 1 k + a + 1 k + c + 1 k + c &gt; 3 k + 1 So one would hope that the k = 3 case, lying farther within the "valid" region, might be easier than k = 2 but I have not been able to prove it.Note that it is not always true, under our constraint, that 1 2 + a + 1 2 + c + 1 2 + c ≥ 1 3 + a + 1 3 + c + 1 3 + c + 1 4 , which if true would prove the k = 3 case immediately.

Question

Show that if       1    a    +      1    b    +      1    c    =  a  +  b  +  c, then       1          3      +      a        +      1          3      +      c        +      1          3      +      c        ≤      3    4  The corresponding problem replacing the   3s with   2 is shown here:How to prove       1          2      +      a        +      1          2      +      b        +      1          2      +      c        ≤  1?The proof is not staggeringly difficult: The idea is to show (the tough part) that if       1          2      +      a        +      1          2      +      c        +      1          2      +      c        =  1 then       1    a    +      1    b    +      1    c    ≥  a  +  b  +  c. The result then easily follows.A pair of observations, both under the constraint of       1    a    +      1    b    +      1    c    =  a  +  b  +  c:If   k  ≥  2 then       1          k      +      a        +      1          k      +      c        +      1          k      +      c        ≤      3          k      +      1      . This is proven for   k  =  2 but I don&#039;t see how for   k  &amp;gt;  2If   0  &amp;lt;  k  &amp;lt;  2 then there are positive   (  a  ,  b  ,  c  ) satisfying the constraint such that       1          k      +      a        +      1          k      +      c        +      1          k      +      c        &amp;gt;      3          k      +      1      So one would hope that the   k  =  3 case, lying farther within the &quot;valid&quot; region, might be easier than   k  =  2 but I have not been able to prove it.Note that it is not always true, under our constraint, that       1          2      +      a        +      1          2      +      c        +      1          2      +      c        ≥      1          3      +      a        +      1          3      +      c        +      1          3      +      c        +      1    4  , which if true would prove the   k  =  3 case immediately.

Sarahi Thomas · Accepted Answer

By C-S      ∑          c      y      c            1          3      +      a        ≤      1          (      3      +      1              )        2                  ∑          c      y      c            (                  3        2                    2        +        a              +                  1        2            1        )    =  =      9    16        ∑          c      y      c            1          2      +      a        +      3    16    ≤      9    16    +      3    16    =      3    4  and we are done!

Rocco Juarez · Accepted Answer

∑                  cyc                  1          2      +      a        ≤  1Use (C-S)  ∑      (                  α        i        2                    x        i              )    ≥                    (        ∑                  α          i                )            2              ∑              x        i            for the   n  =  2 case with                              α          1                =        3                              α          2                =        k        −        2                                      x          1                =        a        +        2                              x          2                =        k        −        2            giving            3      2              2      +      a        +            (      k      −      2              )        2                    k      −      2        ≥            (      k      +      1              )        2                    k      +      a      Similarly for   b in place of   a, and for   c in place of   a. Adding the   a,   b and   c inequalities we get      ∑                  cyc                  1          k      +      a        ≤      1          (      k      +      1              )        2                  ∑                  cyc                  (                  3        2                    2        +        a              +    (    k    −    2    )    )    =  3            (      k      −      2      )              (      k      +      1              )        2              +      9          (      k      +      1              )        2                  ∑                  cyc                        3      2              2      +      a      Now we use our given theorem to write      ∑                  cyc                  1          k      +      a        ≤  3            (      k      −      2      )              (      k      +      1              )        2              +      9          (      k      +      1              )        2              =            3      k      +      3              (      k      +      1              )        2              =      3          k      +      1

Show that if 1/a+1/b+1/c=a+b+c, then 1/(3+a)+1/(3+c)+1/(3+c)<=3/4

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