Is it true that a a + b + b b + c + c c + a ≥ 3 2 if a + b + c = 1?

Question

Is it true that       a          a      +      b        +      b          b      +      c        +      c          c      +      a        ≥      3    2   if   a  +  b  +  c  =  1?

Miya Swanson · Accepted Answer

I think it&#039;s wrong.Try   c  →      0    +  .We obtain      a          a      +      b        ≥      1    2  or  a  ≥  band you can get a counterexample with   a  &amp;lt;  bAlso, we can see it for   a  →      0    +  : we obtain   0  ≥      1    2    .By the way, your second problem is true.Indeed, we need to prove that      ∑          c      y      c                          a        2            +      b              b      +      c        ≥  2or      ∑          c      y      c        (      a    2    +  b  )  (  a  +  b  )  (  a  +  c  )  ≥  2      ∏          c      y      c        (  a  +  b  )or      ∑          c      y      c        (      a    2    +  b  )  (      a    2    +  a  b  +  a  c  +  b  c  )  ≥  2      ∏          c      y      c        (  a  +  b  )or      ∑          c      y      c        (      a    2    +  b  )  (  a  +  b  c  )  ≥  2      ∏          c      y      c        (  a  +  b  )or      ∑          c      y      c            (          a      3        +          1      3        a    b    c    +    a    b    +          a      2        b    )    ≥  2      ∑          c      y      c            (          a      2        b    +          a      2        c    +          2      3        a    b    c    )  or      ∑          c      y      c            (          a      3        +          1      3        a    b    c    +          a      2        b    +          a      2        c    +    a    b    c    +          a      2        b    )    ≥  2      ∑          c      y      c            (          a      2        b    +          a      2        c    +          2      3        a    b    c    )  or      ∑          c      y      c        (      a    3    −      a    2    c  )  ≥  0  ,which is true by Rearrangement.

kjukks1234531 · Accepted Answer

Computing the left hand side minus the right hand side we obtain

1 / 2 \frac{(b - c) (a - c) (a - b)}{(a + b) (b + c) (c + a)} \geq 0

Is it true that a/(a+b)+b/(b+c)+c/(c+a)>= 3/2 if a+b+c=1?

Answered question

Answer & Explanation

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