For a , b , c &lt; 1. Minimize M = a 2 ( 1 − 2 b ) b + b 2 ( 1 − 2 c ) c + c 2 ( 1 − 2 a ) a My Try: From a b + b c + c a = 1 ⇔ 3 ≤ a + b + c &lt; 3We have: M = a 2 b + b 2 c + c 2 a − 2 ( a 2 + b 2 + c 2 ) ≥ ( a + b + c ) 2 a + b + c − 2 ( a 2 + b 2 + c 2 ) − 4 ( a b + b c + c a ) + 4 ( a b + b c + c a ) = a + b + c − 2 ( a + b + c ) 2 + 4 ( 1 )Let a + b + c = x ( 3 ≤ x &lt; 3 )We find Min of function y = − 2 x 2 + x + 4And y M i n = − 2 + 3 when x = 3 Right or Wrong ?

Question

For   a  ,  b  ,  c  &amp;lt;  1. Minimize   M  =                    a        2                    (        1        −        2        b        )              b    +                    b        2                    (        1        −        2        c        )              c    +                    c        2                    (        1        −        2        a        )              a  My Try: From   a  b  +  b  c  +  c  a  =  1  ⇔      3    ≤  a  +  b  +  c  &amp;lt;  3We have:   M  =            a      2        b    +            b      2        c    +            c      2        a    −  2      (          a      2        +          b      2        +          c      2        )    ≥                    (        a        +        b        +        c        )            2              a      +      b      +      c        −  2      (          a      2        +          b      2        +          c      2        )    −  4      (    a    b    +    b    c    +    c    a    )    +  4      (    a    b    +    b    c    +    c    a    )    =  a  +  b  +  c  −  2  (  a  +  b  +  c      )    2    +  4  (  1  )Let   a  +  b  +  c  =  x  (      3    ≤  x  &amp;lt;  3  )We find Min of function   y  =  −  2      x    2    +  x  +  4And       y          M      i      n        =  −  2  +      3   when   x  =      3  Right or Wrong ?

jbacapzh · Accepted Answer

Let   a  =  b  =  c  =      1          3      . Hence,   M  =      3    −  2We&#039;ll prove that it&#039;s a minimal value.Indeed, Let   c  =  min  {  a  ,  b  ,  c  }. Hence, we need to prove that            a      2        b    +            b      2        c    +            c      2        a    −            2      (              a        2            +              b        2            +              c        2            )              a      b      +      a      c      +      b      c        ≥  (      3    −  2  )      a    b    +    a    c    +    b    c  or      a    b    +    a    c    +    b    c        (                  a        2            b        +                  b        2            c        +                  c        2            a        −          3      (      a      b      +      a      c      +      b      c      )        )    ≥  2  (      a    2    +      b    2    +      c    2    −  a  b  −  a  c  −  b  c  )or            a      2        b    +            b      2        a    −  a  −  b  +            b      2        c    +            c      2        a    −            b      2        a    −  c  +  a  +  b  +  c  −      3    (    a    b    +    a    c    +    b    c    )    ≥  ≥  2  (  (  a  −  b      )    2    +  (  c  −  a  )  (  c  −  b  )  )or            (      a      −      b              )        2            (      a      +      b      )              a      b        +            (      c      −      a      )      (              c        2            −              b        2            )              a      c        +  a  +  b  +  c  −      3    (    a    b    +    a    c    +    b    c    )    ≥  ≥  2  (  (  a  −  b      )    2    +  (  c  −  a  )  (  c  −  b  )  )or  (  a  −  b      )    2        (          1      a        +          1      b        −    2    )    +  (  c  −  a  )  (  c  −  b  )      (          1      a        +          b              a        c              −    2    )    +  a  +  b  +  c  −      3    (    a    b    +    a    c    +    b    c    )    ≥  0  ,which is true because       1    a    +      1    b    −  2  &amp;gt;  0,       1    a    +      b          a      c        −  2  ≥      1    a    +      1    a    −  2  &amp;gt;  0 and  a  +  b  +  c  −      3    (    a    b    +    a    c    +    b    c    )    =            (      a      −      b              )        2            +      (      c      −      a      )      (      c      −      b      )              a      +      b      +      c      +              3        (        a        b        +        a        c        +        b        c        )              ≥  0.Done!The inequality            a      2        b    +            b      2        c    +            c      2        a    −            2      (              a        2            +              b        2            +              c        2            )              a      b      +      a      c      +      b      c        ≥  (      3    −  2  )      a    b    +    a    c    +    b    c  is true for all positives   a,   b and   c, but my proof of the last statement is very ugly.

For 0<a,b,c<1 and ab+bc+ca=1, minimize M=(a^2(1-2b))/(b)+(b^2(1-2c))/(c)+(c^2(1-2a))/(a)

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Answer & Explanation

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