Prove that a a + b + b b + c + c c + a ≥ 3 2 For a ≥ b ≥ c &gt; 0. Prove that a = 100 ; b = 1 ; c = 1 / 100 it's wrong ??? a a + b + b b + c + c c + a ≥ 3 2 &lt;=&gt; ∑ a a + b − 1 2 + b b + c − 1 2 + c c + a − 1 2 ≥ 0 &lt;=&gt; ∑ a − b 2 ( a + b ) ≥ 0 &lt;=&gt; x y 2 + y z 2 + x z 2 − x 2 y − y 2 z − z 2 x ≥ 0 &lt;=&gt; ( y − x ) ( z − y ) ( z − x ) ≥ 0And we have y − x ≤ 0 ; z − x ≤ 0 ; z − y ≥ 0we are done !?

Question

Prove that       a          a      +      b        +      b          b      +      c        +      c          c      +      a        ≥      3    2  For   a  ≥  b  ≥  c  &amp;gt;  0. Prove that  a  =  100  ;  b  =  1  ;  c  =  1      /    100 it&#039;s wrong ???            a              a        +        b              +            b              b        +        c              +            c              c        +        a              ≥            3      2        &amp;lt;=&amp;gt;  ∑      a          a      +      b        −      1    2    +      b          b      +      c        −      1    2    +      c          c      +      a        −      1    2    ≥  0  &amp;lt;=&amp;gt;  ∑            a      −      b              2      (      a      +      b      )        ≥  0  &amp;lt;=&amp;gt;  x      y    2    +  y      z    2    +  x      z    2    −      x    2    y  −      y    2    z  −      z    2    x  ≥  0  &amp;lt;=&amp;gt;  (  y  −  x  )  (  z  −  y  )  (  z  −  x  )  ≥  0And we have   y  −  x  ≤  0  ;  z  −  x  ≤  0  ;  z  −  y  ≥  0we are done !?

jbaekk7q · Accepted Answer

For   a  ≥  b  ≥  c  &amp;gt;  0 we obtain:      ∑          c      y      c            a          a      +      b        −      3    2    =      ∑          c      y      c            (          a              a        +        b              −          1      2        )    =      ∑          c      y      c                  a      −      b              2      (      a      +      b      )        =  =      ∑          c      y      c                  (      a      −      b      )      (              c        2            +      a      b      +      a      c      +      b      c      )              2              ∏                  c          y          c                    (      a      +      b      )        =      ∑          c      y      c                          c        2            a      −              c        2            b              2              ∏                  c          y          c                    (      a      +      b      )        =  =      ∑          c      y      c                          a        2            b      −              a        2            c              2              ∏                  c          y          c                    (      a      +      b      )        =            (      a      −      b      )      (      a      −      c      )      (      b      −      c      )              2              ∏                  c          y          c                    (      a      +      b      )        ≥  0

For a>=b>=c>0. Prove that a/(a+b)+b/(b+c)+c/(c+a)>=3/2 a=100;b=1;c=1/100 it's wrong ???

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