3 a b c ≥ a + b + c, prove that 1 a + 1 b + 1 c ≥ a + b + c

Question

3          a      b      c        ≥  a  +  b  +  c, prove that       1    a    +      1    b    +      1    c    ≥  a  +  b  +  c

crearti2d4 · Accepted Answer

(                        1          a                    +                        1          b                    +                        1          c                    )        2    ≥  3      (                  1                  a          b                      +                  1                  b          c                      +                  1                  c          a                      )    ≥  a  b  c  (  a  +  b  +  c  )      (                  1                  a          b                      +                  1                  b          c                      +                  1                  c          a                      )    =  (  a  +  b  +  c      )    2

Madelynn Winters · Accepted Answer

The condition gives  1  ≥                    (        a        +        b        +        c        )        a        b        c            3        .Thus,      1    a    +      1    b    +      1    c    ≥      (          1      a        +          1      b        +          1      c        )                      (        a        +        b        +        c        )        a        b        c            3        =  =                    (        a        b        +        a        c        +        b        c                  )          2                (        a        +        b        +        c        )                    3        a        b        c              ≥                    3        a        b        c        (        a        +        b        +        c        )        (        a        +        b        +        c        )                    3        a        b        c              =  a  +  b  +  c  .

(3)/(abc) >= a + b + c, prove that 1/a + 1/b + 1/c >= a + b + c

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