Prove this inequality ( a + b + c ) ( b + c + d ) 3 ≥ a b 3 + c d 3 Let a , b , c , d &gt; 0Idea: Hence, we need to prove that ( a + b + c ) ( b + c + d ) ≥ a b + c d + 3 ( a b ) 2 c d 3 + 3 ( c d ) 2 ( a b ) 3 ⟺ a c + b c + b d + a d + b 2 + c 2 ≥ 3 ( a b ) 2 c d 3 + 3 ( c d ) 2 ( a b ) 3 I attempted a following proof by AM-GM inequality,But I am not able to solve the upper part, this inequality

Question

Prove this inequality              (      a      +      b      +      c      )      (      b      +      c      +      d      )        3    ≥            a      b        3    +            c      d        3  Let   a  ,  b  ,  c  ,  d  &amp;gt;  0Idea: Hence, we need to prove that  (  a  +  b  +  c  )  (  b  +  c  +  d  )  ≥  a  b  +  c  d  +  3            (      a      b              )        2            c      d        3    +  3            (      c      d              )        2            (      a      b      )        3    ⟺  a  c  +  b  c  +  b  d  +  a  d  +      b    2    +      c    2    ≥  3            (      a      b              )        2            c      d        3    +  3            (      c      d              )        2            (      a      b      )        3  I attempted a following proof by AM-GM inequality,But I am not able to solve the upper part, this inequality

Kaleb Harrell · Accepted Answer

It&#039;s just Holder:  (  a  +  b  +  c  )  (  b  +  c  +  d  )      (          b              b        +        c              +          c              b        +        c              )    ≥  ≥            (                        a          ⋅          (          b          +          c          )          ⋅                      b                          b              +              c                                      3            +                        (          b          +          c          )          ⋅          d          ⋅                      c                          b              +              c                                      3            )        3    =            (                        a          b                3            +                        c          d                3            )        3

Let a,b,c,d>0 show that root[3]((a+b+c)(b+c+d))>=root[3](ab)+root[3](cd)

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