Inequality 3 1 1 + a + 1 1 + b + 1 1 + c ≥ 1 + 3 1 a + 1 b + 1 c for positive aMyTry::We can write A and B as = 1 1 + a + 1 1 + b + 1 1 + c = 3 A and 1 a + 1 b + 1 c = 3 B Using cauchyschwarz Inequality 1 1 + a + 1 1 + b + 1 1 + c ≥ 3 2 1 + a + 1 + b + 1 + c = 9 3 + a + b + c Now How can i solve after that , Help Required, Thanks

Question

Inequality       3                  1                  1          +          a                    +              1                  1          +          b                    +              1                  1          +          c                      ≥  1  +      3                  1        a            +              1        b            +              1        c             for positive   aMyTry::We can write   A and   B as  =      1          1      +      a        +      1          1      +      b        +      1          1      +      c        =      3    A  and      1    a    +      1    b    +      1    c    =      3    B  Using cauchyschwarz Inequality      1          1      +      a        +      1          1      +      b        +      1          1      +      c        ≥            3      2              1      +      a      +      1      +      b      +      1      +      c        =      9          3      +      a      +      b      +      c      Now How can i solve after that , Help Required, Thanks

Anika Stevenson · Accepted Answer

we need to prove that            3      (      1      +      a      )      (      1      +      b      )      (      1      +      c      )                      ∑                  c          y          c                    (      a      b      +      2      a      +      1      )        ≥  1  +            3      a      b      c              a      b      +      a      c      +      b      c      or      ∑          c      y      c        (  2      a    2        b    2    −  2      a    2    b  c  +      a    2    b  +      a    2    c  −  2  a  b  c  )  ≥  0or      ∑          c      y      c        (  a  −  b      )    2    (      c    2    +  c  )  ≥  0Done!

Izabella Ponce · Accepted Answer

More way.We&#039;ll rewrite our inequality in the following form  3      ∑          c      y      c            1          a      (      1      +      a      )        ≥      ∑          c      y      c            1          1      +      a            ∑          c      y      c            1    a  which is Rearrangement.

Inequality 3 1 1 + a

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