Any ideas for this cyclic inequality?If a , b , c , d &gt; 0 and a + b + c + d = 8, prove that a b 2 + 2 b + b c 2 + 2 c + c d 2 + 2 d + d a 2 + 2 a ⩾ 16 ( a + c ) ( b + d ) .

Question

Any ideas for this cyclic inequality?If   a  ,  b  ,  c  ,  d  &amp;gt;  0 and   a  +  b  +  c  +  d  =  8, prove that      a                  b        2            +      2      b        +      b                  c        2            +      2      c        +      c                  d        2            +      2      d        +      d                  a        2            +      2      a        ⩾      16          (      a      +      c      )      (      b      +      d      )        .

Zoey Benitez · Accepted Answer

By Holder and AM-GM we obtain:      ∑          c      y      c            a                  b        2            +      2      b        =                    ∑                  c          y          c                    a      b              ∑                  c          y          c                    a      (      b      +      2      )              ∑                  c          y          c                            a                  b          (          b          +          2          )                                    ∑                  c          y          c                    a      b              ∑                  c          y          c                    a      (      b      +      2      )        ≥            (      a      +      b      +      c      +      d              )        3                    (      a      +      c      )      (      b      +      d      )              ∑                  c          y          c                    (      a      b      +      2      a      )        =  =      512          (      a      +      c      )      (      b      +      d      )      (      (      a      +      c      )      (      b      +      d      )      +      16      )        ≥      512          (      a      +      c      )      (      b      +      d      )              (                              (                                          a                +                c                +                b                +                d                            2                        )                    2                +        16        )              =  =      16          (      a      +      c      )      (      b      +      d      )        .

Any ideas for this cyclic inequality? If a,b,c,d>0 and a+b+c+d=8, prove that a/(b^2+2b)+ b/(c^2+2c)+c/(d^2+2d)+\d/(a^2+2a)>=(16)/((a+c)(b+d)).

Answered question

Answer & Explanation

New Questions in Pre-Algebra