Inequality b a b + b + 1 + c b c + c + 1 + a a c + a + 1 ≥ 3 m m 2 + m + 1 Let m = ( a b c ) 1 3 , where a , b , c ∈ R + . Then prove that b a b + b + 1 + c b c + c + 1 + a a c + a + 1 ≥ 3 m m 2 + m + 1 In this inequality I first applied Titu's lemma ; then Rhs will come 9/(some terms) ; now to maximise the rhs I tried to minimise the denominator by applying AM-GM inequality.But then the reverse inequality is coming Please help.

Question

Inequality       b          a      b      +      b      +      1        +      c          b      c      +      c      +      1        +      a          a      c      +      a      +      1        ≥            3      m                      m        2            +      m      +      1      Let   m  =  (  a  b  c      )                  1        3            , where   a  ,  b  ,  c  ∈            R              +            . Then prove that      b          a      b      +      b      +      1        +      c          b      c      +      c      +      1        +      a          a      c      +      a      +      1        ≥            3      m                      m        2            +      m      +      1      In this inequality I first applied Titu&#039;s lemma ; then Rhs will come 9/(some terms) ; now to maximise the rhs I tried to minimise the denominator by applying AM-GM inequality.But then the reverse inequality is coming Please help.

edytorialkp · Accepted Answer

By Holder:      ∑          c      y      c            b          a      b      +      b      +      1        =  1  −            (      a      b      c      −      1              )        2                            ∏                  c          y          c                    (      a      b      +      b      +      1      )        ≥  1  −            (              m        3            −      1              )        2                    (                                    a            2                                b            2                                c            2                          3            +                        a          b          c                3            +      1              )        3              =  =  1  −            (      m      −      1              )        2            (              m        2            +      m      +      1              )        2                    (              m        2            +      m      +      1              )        3              =  1  −            (      m      −      1              )        2                            m        2            +      m      +      1        =            3      m                      m        2            +      m      +      1        .Done!

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