Prove inequality ∑ a 3 − b 3 ( a − b ) 3 ≥ 9 4 Given a , b , c are positive number. Prove that a 3 − b 3 ( a − b ) 3 + b 3 − c 3 ( b − c ) 3 + c 3 − a 3 ( c − a ) 3 ≥ 9 4 ⇔ ∑ 3 ( a + b ) 2 + ( a − b ) 2 ( a − b ) 2 ≥ 9 ⇔ ( a + b ) 2 ( a − b ) 2 + ( b + c ) 2 ( b − c ) 2 + ( c + a ) 2 ( c − a ) 2 ≥ 2Which a + b a − b . b + c b − c + b + c b − c . c + a c − a + c + a c − a . a + b a − b = − 1Use x 2 + y 2 + z 2 ≥ − 2 ( x y + y z + z x ) then inequality rightI don't know why use x 2 + y 2 + z 2 ≥ − 2 ( x y + y z + z x ) then inequality right?P/s: Sorry i knowed, let x = a + b a − b We have ( x + 1 ) ( y + 1 ) ( z + 1 ) = ( x − 1 ) ( y − 1 ) ( z − 1 ) =&gt; x y + y z + z x = − 1

Question

Prove inequality   ∑                              a          3                    −                        b          3                                                                        (                              a                −                b                            )                                3                      ≥      9    4  Given   a  ,  b  ,  c are positive number. Prove that                    a        3            −              b        3                            (        a        −        b        )            3        +                    b        3            −              c        3                            (        b        −        c        )            3        +                    c        3            −              a        3                            (        c        −        a        )            3        ≥      9    4    ⇔  ∑            3      (      a      +      b              )        2            +      (      a      −      b              )        2                    (      a      −      b              )        2              ≥  9  ⇔            (      a      +      b              )        2                    (      a      −      b              )        2              +            (      b      +      c              )        2                    (      b      −      c              )        2              +            (      c      +      a              )        2                    (      c      −      a              )        2              ≥  2Which            a      +      b              a      −      b        .            b      +      c              b      −      c        +            b      +      c              b      −      c        .            c      +      a              c      −      a        +            c      +      a              c      −      a        .            a      +      b              a      −      b        =  −  1Use       x    2    +      y    2    +      z    2    ≥  −  2  (  x  y  +  y  z  +  z  x  ) then inequality rightI don&#039;t know why use       x    2    +      y    2    +      z    2    ≥  −  2  (  x  y  +  y  z  +  z  x  ) then inequality right?P/s: Sorry i knowed, let   x  =            a      +      b              a      −      b      We have   (  x  +  1  )  (  y  +  1  )  (  z  +  1  )  =  (  x  −  1  )  (  y  −  1  )  (  z  −  1  )  =&amp;gt;  x  y  +  y  z  +  z  x  =  −  1

Coraducci0d · Accepted Answer

I think it means that the variables are reals.By your identity we obtain            (              ∑                  c          y          c                                      a          +          b                          a          −          b                    )        2    =      ∑          c      y      c                  (      a      +      b              )        2                    (      a      −      b              )        2              −  2  ≥  0Thus,      ∑          c      y      c                  (      a      +      b              )        2                    (      a      −      b              )        2              ≥  2or      ∑          c      y      c            (                  (        a        +        b                  )          2                            (        a        −        b                  )          2                      −    1    )    ≥  −  1or      ∑          c      y      c                  4      a      b              (      a      −      b              )        2              ≥  −  1or      ∑          c      y      c                  3      a      b              (      a      −      b              )        2              ≥  −      3    4  or      ∑          c      y      c            (                  3        a        b                    (        a        −        b                  )          2                      +    1    )    ≥  3  −      3    4  or      ∑          c      y      c                          a        2            +      a      b      +              b        2                    (      a      −      b              )        2              ≥      9    4  or      ∑          c      y      c                          a        3            +              b        3                    (      a      −      b              )        3              ≥      9    4    .Done!

monopolutk · Accepted Answer

(  x  +  y  +  z      )    2    ≥  0        x    2    +      y    2    +      z    2    +  2  x  y  +  2  x  z  +  2  y  z  ≥  0        x    2    +      y    2    +      z    2    ≥  −  2  (  x  y  +  y  z  +  z  x  )

Given a,b,c are positive number. Prove that (a^3-b^3)/((a-b)^3)+(b^3-c^3)/((b-c)^3}+(c^3-a^3)/((c-a)^3)>=(9)/(4)

Open question

Answer & Explanation

New Questions in Pre-Algebra