Find the minimum of ∑ c y c a 2 ( b + c ) with a , b , c &gt; 0As said in the title, I have to find the minimum of the following: ∑ c y c a 2 ( b + c ) with a + b + c &gt; 0In my very last attempt, I tried to work it out using AM-GM: Since a 2 ( b + c ) = 1 2 a a ( b + c ) ≥ 2 a a + b + c ⇒ ∑ c y c a 2 ( b + c ) ≥ 2 ( a + b + c ) a + b + c = 2 The problem is, to get the minimum value, a = b + c b = c + a c + a + bor a = b = c = 0. This is wrong since a , b , c &gt; 0 And after some calculus I found the minimum value is 3 2 when a = b = c. So what mistake I have made?

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Find the minimum of       ∑          c      y      c                          a                  2          (          b          +          c          )                     with   a  ,  b  ,  c  &amp;gt;  0As said in the title, I have to find the minimum of the following:      ∑          c      y      c                          a                  2          (          b          +          c          )                    with   a  +  b  +  c  &amp;gt;  0In my very last attempt, I tried to work it out using AM-GM: Since            a              2        (        b        +        c        )              =      1          2            a          a      (      b      +      c      )        ≥                    2            a              a      +      b      +      c        ⇒      ∑          c      y      c                          a                  2          (          b          +          c          )                      ≥                    2            (      a      +      b      +      c      )              a      +      b      +      c        =      2  The problem is, to get the minimum value,  a  =  b  +  c  b  =  c  +  a  c  +  a  +  bor   a  =  b  =  c  =  0. This is wrong since   a  ,  b  ,  c  &amp;gt;  0 And after some calculus I found the minimum value is       3    2   when   a  =  b  =  c. So what mistake I have made?

fishsnortgth4o · Accepted Answer

By AM-GM       ∑          c      y      c                  a              2        (        b        +        c        )              =      ∑          c      y      c                          2            a              2              a        (        b        +        c        )              ≥      ∑          c      y      c                          2            a              a      +      b      +      c        =      2  , which is infimum because we can use   c  →      0    +   and   b  =  c

Martin Nunez · Accepted Answer

Stronger inequality: With   a  ,  b  ,  c  &amp;gt;  0            a              b        +        c              +            b              c        +        a              +            c              a        +        b              ≥      2    +    2.                  a        b        c                              (          a          +          b          )                          (          b          +          c          )                          (          c          +          a          )                    Proof: Thus, we have:      (    a    +    b    )        (    b    +    c    )        (    c    +    a    )    +  a  b  c  =      (    a    +    b    +    c    )        (    a    b    +    b    c    +    c    a    )  From the assumed inequality, we obtain the equivalent inequality:      a          (      a      +      b      )              (      a      +      c      )        +      b          (      b      +      c      )              (      b      +      a      )        +      c          (      c      +      a      )              (      c      +      b      )        ≥  2            (      a      +      b      +      c      )              (      a      b      +      b      c      +      c      a      )      Other way, we have:      a          (      a      +      b      )              (      a      +      c      )        +      b          (      b      +      c      )              (      b      +      a      )        +      c          (      c      +      a      )              (      c      +      b      )        =            a              2                    (      a      +      b      +      c      )        +    a    b    c    +            b              2                    (      a      +      b      +      c      )        +    a    b    c    +            c              2                    (      a      +      b      +      c      )        +    a    b    c    ≥                    (        a                  a          +          b          +          c                +        b                  a          +          b          +          c                +        c                  a          +          b          +          c                )                    2              +    9    a    b    c    =                    (        a        +        b        +        c        )                    3              +    9    a    b    c    ≥  4      (    a    +    b    +    c    )        (    a    b    +    b    c    +    c    a    )  Done!

Find the minimum of <munder> <mo movablelimits="false">∑ <mrow class=

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