Geometric (Trigonometric) inequality ( a + b + c ) 3 3 a b c ≤ 1 + 4 R r

Question

Geometric (Trigonometric) inequality             (      a      +      b      +      c              )        3                    3      a      b      c        ≤  1  +            4      R        r

Raven Higgins · Accepted Answer

We can prove that            (      a      +      b      +      c              )        3                    3      a      b      c        ≤  1  +            4      R        r  Indeed, we need to prove that            (      a      +      b      +      c              )        3                    3      a      b      c        ≤  1  +                    a        b        c            S                      2        S                    a        +        b        +        c            or            (      a      +      b      +      c              )        3                    3      a      b      c        ≤  1  +            8      a      b      c      (      a      +      b      +      c      )              16              S        2            or            (      a      +      b      +      c              )        3                    3      a      b      c        ≤  1  +            8      a      b      c                      ∏                  c          y          c                    (      a      +      b      −      c      )      or            (      a      +      b      +      c              )        3                    3      a      b      c        ≤                    ∑                  c          y          c                    (      −              a        3            +              a        2            b      +              a        2            c      +      2      a      b      c      )                      ∏                  c          y          c                    (      a      +      b      −      c      )      or            (      a      +      b      +      c              )        3                    3      a      b      c        ≤                    ∑                  c          y          c                    (      −              a        3            +      a      b      c      +              a        2            b      +              a        2            c      +      a      b      c      )                      ∏                  c          y          c                    (      a      +      b      −      c      )                  (      a      +      b      +      c              )        2                    3      a      b      c        ≤                    ∑                  c          y          c                    (      −              a        2            +      a      b      +      a      b      )                      ∏                  c          y          c                    (      a      +      b      −      c      )      or      ∑          c      y      c        a      ∑          c      y      c        (  2      a    2        b    2    −      a    4    )  ≤  3      ∑          c      y      c        (  2      a    2        b    2    c  −      a    3    b  c  )or  2      ∑          c      y      c        (      a    4    b  +      a    4    c  −      a    3        b    2    −      a    3        c    2    −  2      a    3    b  c  +  2      a    2        b    2    c  )  +      ∑          c      y      c        (      a    5    −      a    4    b  −      a    4    c  +      a    3    b  c  )  ≥  0  ,for which it&#039;s enough to prove that      ∑          c      y      c        (  a  −  b      )    2    (  a  b  (  a  +  b  )  −  a  b  c  )  ≥  0or      ∑          c      y      c        (  a  −  b      )    2    a  b  (  a  +  b  −  c  )  ≥  0.Done!

Geometric (Trigonometric) inequality ( a + b +

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