Showing 64 ≤ ( 1 + 1 x ) ( 1 + 1 y ) ( 1 + 1 z ) subject to x , y , z &gt; 0 and x + y + z = 1This inequality is equivalent to; ( x + 1 4 ⋅ 1 x ) ( y + 1 4 ⋅ 1 y ) ( z + 1 4 ⋅ 1 z ) ≥ 1 taking logarithm of both sides we have to show that LHS is greater or equal to 0, but by convexity of ln ⁡ ( x + 1 4 ) − ln ⁡ ( x ), i.e. d 2 d x 2 [ ln ⁡ ( x + 1 4 ) − ln ⁡ ( x ) ] = 1 x 2 − 1 ( x + 1 ) 2 &gt; 0 we have ∑ c y c ln ⁡ ( x + 1 4 ) − ln ⁡ ( x ) ≥ 3 ln ⁡ ( ∑ c y c x + 1 4 ) − 3 ln ⁡ ( ∑ c y c x ) = 3 ln ⁡ ( 1 ) − 3 ln ⁡ ( 1 ) = 0Is this OK, or is there a much simpler way ?

Question

Showing   64  ≤      (    1    +          1      x        )        (    1    +          1      y        )        (    1    +          1      z        )   subject to   x  ,  y  ,  z  &amp;gt;  0 and   x  +  y  +  z  =  1This inequality is equivalent to;      (                  x        +        1            4        ⋅          1      x        )        (                  y        +        1            4        ⋅          1      y        )        (                  z        +        1            4        ⋅          1      z        )    ≥  1 taking logarithm of both sides we have to show that LHS is greater or equal to   0, but by convexity of   ln  ⁡      (                  x        +        1            4        )    −  ln  ⁡  (  x  ), i.e.             d      2              d              x        2                  [    ln    ⁡          (                        x          +          1                4            )        −    ln    ⁡    (    x    )    ]    =      1          x      2        −      1          (      x      +      1              )        2              &amp;gt;  0 we have      ∑          c      y      c        ln  ⁡      (                  x        +        1            4        )    −  ln  ⁡  (  x  )  ≥  3  ln  ⁡      (          ∑              c        y        c                            x        +        1            4        )    −  3  ln  ⁡      (          ∑              c        y        c              x    )    =  3  ln  ⁡  (  1  )  −  3  ln  ⁡  (  1  )  =  0Is this OK, or is there a much simpler way ?

Holly Crane · Accepted Answer

By AM-GM we have:      1    3    =      1    3    (  x  +  y  +  z  )  ≥            x      y      z        3      ⟹        1    27    ≥  x  y  z    ⟹        1          27      x      y      z        ≥  1.Using AM-GM again, with 4 terms this time, we obtain  1  +      1    x    =  1  +      1          3      x        +      1          3      x        +      1          3      x        ≥  4            1              27                  x          3                      4    .Do that for the other 2 terms and multiply. We get      (    1    +          1      x        )        (    1    +          1      y        )        (    1    +          1      z        )    ≥  64                    (                  1                      27            x            y            z                          )            3        4    ≥  64.Equalities are iff   x  =  y  =  z  =      1    3

Yair Valentine · Accepted Answer

By Holder and AM-GM       ∏          c      y      c            (    1    +          1      x        )    ≥            (      1      +              1                              x            y            z                    3                    )        3    ≥            (      1      +              1                  1          3                    )        3    =  64

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