Does ( x + 1 ) log ⁡ ( x + 1 ) − 2 x log ⁡ x + ( x − 1 ) log ⁡ ( x − 1 ) ≥ 1 x hold for x ≥ 1?While calculating some integrals I happened to face the following estimate: ∫ m m + 1 ∫ n n + 1 d y d x x + y ≥ 1 m + n + 1 .After some tedious calculations, I figured that this estimate follows from the inequality ( x + 1 ) log ⁡ ( x + 1 ) − 2 x log ⁡ x + ( x − 1 ) log ⁡ ( x − 1 ) ≥ 1 x for x &gt; 1.(If we interpret 0 log ⁡ 0 as lim ϵ → 0 ϵ log ⁡ ϵ = 0, then the inequality also holds for x = 1.)But how do we prove this inequality?What I have tried: Let f ( x ) = x log ⁡ x − ( x − 1 ) log ⁡ ( x − 1 ) for x &gt; 1. Then the RHS equals f ( x + 1 ) − f ( x ) so by the Mean Value Theorem there exist some ξ between x and x + 1 such that f ( x + 1 ) − f ( x ) = f ′ ( ξ ) = log ⁡ ( 1 + 1 ξ − 1 ) ,and it suffices to show that log ⁡ ( 1 + 1 x ) ≥ 1 x ... which is unfortunately not valid !I think some clever use of the MVT can solve this problem, but I don't see how I should proceed. Please enlighten me.

Question

Does   (  x  +  1  )  log  ⁡      (    x    +    1    )    −  2  x  log  ⁡      x    +  (  x  −  1  )  log  ⁡      (    x    −    1    )    ≥      1    x   hold for   x  ≥  1?While calculating some integrals I happened to face the following estimate:      ∫    m          m      +      1            ∫    n          n      +      1                  d      y      d      x              x      +      y        ≥      1          m      +      n      +      1        .After some tedious calculations, I figured that this estimate follows from the inequality  (  x  +  1  )  log  ⁡      (    x    +    1    )    −  2  x  log  ⁡      x    +  (  x  −  1  )  log  ⁡      (    x    −    1    )    ≥      1    x      for   x  &amp;gt;  1.(If we interpret   0  log  ⁡  0 as       lim          ϵ      →      0        ϵ  log  ⁡      ϵ    =  0, then the inequality also holds for   x  =  1.)But how do we prove this inequality?What I have tried: Let   f  (  x  )  =  x  log  ⁡      x    −  (  x  −  1  )  log  ⁡      (    x    −    1    )   for   x  &amp;gt;  1. Then the RHS equals   f  (  x  +  1  )  −  f  (  x  ) so by the Mean Value Theorem there exist some   ξ between   x and   x  +  1 such that  f  (  x  +  1  )  −  f  (  x  )  =      f    ′    (  ξ  )  =  log  ⁡            (      1      +              1                  ξ          −          1                    )        ,and it suffices to show that   log  ⁡      (    1    +          1      x        )    ≥      1    x   ... which is unfortunately not valid !I think some clever use of the MVT can solve this problem, but I don&#039;t see how I should proceed. Please enlighten me.

Katelyn Chapman · Accepted Answer

Let   f  (  x  )  =  (  x  +  1  )  ln  ⁡      (    x    +    1    )    −  2  x  ln  ⁡      x    +  (  x  −  1  )  ln  ⁡      (    x    −    1    )    −      1    x  Hence,       f    ″    (  x  )  =      2                  x        3            (              x        2            −      1      )        &amp;gt;  0 for   x  &amp;gt;  1 and the rest is smooth.Because       f    ′    (  x  )  =  ln  ⁡  (  x  −  1  )  −  2  ln  ⁡      x    +  ln  ⁡  (  x  +  1  )  +      1          x      2        &amp;lt;      lim          x      →      +      ∞            f    ′    (  x  )  =  0and   f  (  x  )  &amp;gt;      lim          x      →      +      ∞        f  (  x  )  =      lim          x      →      +      ∞        ln  ⁡                    (        1        +                  1          x                )                    x        +        1                            (        1        +                  1                      x            −            1                          )                    x        −        1              =  ln  ⁡      e    e    =  0

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