Prove that 1 a + b + 1 + 1 b + c + 1 + 1 c + a + 1 ≤ 1If a b c = 1 then 1 a + b + 1 + 1 b + c + 1 + 1 c + a + 1 ≤ 1I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?

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Prove that       1          a      +      b      +      1        +      1          b      +      c      +      1        +      1          c      +      a      +      1        ≤  1If   a  b  c  =  1 then      1          a      +      b      +      1        +      1          b      +      c      +      1        +      1          c      +      a      +      1        ≤  1I have tried AM-GM and C-S and can&#039;t seem to find a solution. What is the best way to prove it?

wespee0 · Accepted Answer

Also we can use the following reasoning.For positives   a,   b and   c let   a  =      x    3  ,   b  =      y    3   and   c  =      z    3  Hence,       ∑          c      y      c            1          a      +      b      +      1        =      ∑          c      y      c            1                  x        3            +              y        3            +      x      y      z        ≤      ∑          c      y      c            1                  x        2            y      +              y        2            x      +      x      y      z        =      ∑          c      y      c            z          x      +      y      +      z        =  1,but I think the first way is better.

Krish Logan · Accepted Answer

It&#039;s wrong, of course.For positives   a,   b and   c it&#039;s       ∑          c      y      c        (      a    2    b  +      a    2    c  −  2  a  )  ≥  0, which is Muirhead because   (  2  ,  1  ,  0  )  ≻      (          5      3        ,          2      3        ,          2      3        )  .

If abc=1 then 1/(a+b+1)+1/(b+c+1)+1/(c+a+1)<=1 I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?

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