a 1 + b 2 c + b 1 + c 2 d + c 1 + d 2 a + d 1 + a 2 b ≥ 2I am stuck in the following olympiad problem:Suppose that a , b , c , d ≥ 0 and a + b + c + d = 4. Prove that a 1 + b 2 c + b 1 + c 2 d + c 1 + d 2 a + d 1 + a 2 b ≥ 2.Attempt. I tried to use reversed AM-GM technique. a 1 + b 2 c − a = − a b 2 c 1 + b 2 c b 1 + c 2 d − b = − b c 2 d 1 + c 2 d c 1 + d 2 a − c = − c d 2 a 1 + d 2 a d 1 + a 2 b − d = − d a 2 b 1 + a 2 b

Question

a          1      +              b        2            c        +      b          1      +              c        2            d        +      c          1      +              d        2            a        +      d          1      +              a        2            b        ≥  2I am stuck in the following olympiad problem:Suppose that   a  ,  b  ,  c  ,  d  ≥  0 and   a  +  b  +  c  +  d  =  4. Prove that      a          1      +              b        2            c        +      b          1      +              c        2            d        +      c          1      +              d        2            a        +      d          1      +              a        2            b        ≥  2.Attempt. I tried to use reversed AM-GM technique.                              a                      1            +                          b              2                        c                          −        a                            =                              −            a                          b              2                        c                                1            +                          b              2                        c                                                        b                      1            +                          c              2                        d                          −        b                            =                              −            b                          c              2                        d                                1            +                          c              2                        d                                                        c                      1            +                          d              2                        a                          −        c                            =                              −            c                          d              2                        a                                1            +                          d              2                        a                                                        d                      1            +                          a              2                        b                          −        d                            =                              −            d                          a              2                        b                                1            +                          a              2                        b

Mohammed Farley · Accepted Answer

By C-S and AM-GM we obtain:      ∑          c      y      c            d          1      +              a        2            b        =      ∑          c      y      c                  d      2              d      +              a        2            b      d        ≥            (      a      +      b      +      c      +      d              )        2                            ∑                  c          y          c                    (      d      +              a        2            b      d      )        =  =      16          4      +      (      a      b      +      c      d      )      (      a      d      +      b      c      )        ≥      16          4      +                        (                                    a              b              +              b              c              +              c              d              +              d              a                        2                    )                2              =  =      16          4      +                        (                                    (              a              +              c              )              (              b              +              d              )                        2                    )                2              ≥      16          4      +                        (                                                    (                                                      a                    +                    c                    +                    b                    +                    d                                    2                                )                            2                        2                    )                2              =  2.

themediamafia73 · Accepted Answer

∑          c      y      c            d          1      +              a        2            b        =      ∑          c      y      c                  d      2              d      +              a        2            b      d      Hence by Titu&#039;s lemma      ∑          c      y      c                  d      2              d      +              a        2            b      d        ≥            (      a      +      b      +      c      +      d              )        2                            ∑                  c          y          c                    (      d      +              a        2            b      d      )        =  =      16          4      +      (      a      b      +      c      d      )      (      a      d      +      b      c      )      Now by AM GM inequality we have            a      b      +      b      c      +      c      d      +      a      d        2    ≥      (    a    b    +    c    d    )    (    a    d    +    b    c    )  Hence we get            (                        a          b          +          b          c          +          c          d          +          a          d                2            )        2    ≥  (  a  b  +  c  d  )  (  a  d  +  b  c  )  ⇒      16          4      +      (      a      b      +      c      d      )      (      a      d      +      b      c      )        ≥      16          4      +                        (                                    a              b              +              b              c              +              c              d              +              a              d                        2                    )                2            Now again by AM GM inequality            a      +      b      +      c      +      d        2    ≥      (    a    +    c    )    (    d    +    b    )    ⇒      16          4      +                        (                                    (              a              +              c              )              (              b              +              d              )                        2                    )                2              ≥      16          4      +                        (                                                    (                                                      a                    +                    b                    +                    c                    +                    d                                    2                                )                            2                        2                    )                2              =  2.

I am stuck in the following olympiad problem: Suppose that a,b,c,d>=0 and a+b+c+d=4. Prove that a/(1+b^2c)+b/(1+c^2d)+c/(1+d^2a)+d/(1+a^2b)>=2

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