Arithmetic Average of Second Chebyshev Function/Lemma 2

Lemma

Let $x \ge 3$ be a real number.

Then:

$-2 \map \ln {x + 1} \le \paren {x - 1} \map \ln {x - 1} - x \ln x$

Proof

Define a function $g : \openint 1 \infty \to \R$ by:

$\map g x = \paren {x - 1} \map \ln {x - 1} - x \ln x + 2 \map \ln {x + 1}$

We have that $g$ is differentiable with:

\(\ds \map {g'} x\)

\(=\)

\(\ds \frac {x - 1} {x - 1} + \map \ln {x - 1} - \frac x x - \ln x + \frac 2 {x + 1}\)

Product Rule for Derivatives, Derivative of Logarithm Function

\(\ds \)

\(=\)

\(\ds \map \ln {x - 1} - \ln x + \frac 2 {x + 1}\)

\(\ds \)

\(=\)

\(\ds \map \ln {1 - \frac 1 x} + \frac 2 {x + 1}\)

\(\ds \)

\(\ge\)

\(\ds \frac 2 {x + 1} + 1 - \frac 1 {1 - \frac 1 x}\)

Bounds of Natural Logarithm

\(\ds \)

\(=\)

\(\ds 1 - \frac x {x - 1} + \frac 2 {x + 1}\)

\(\ds \)

\(=\)

\(\ds \frac {x^2 - 1 - x \paren {x + 1} + 2 \paren {x - 1} } {\paren {x + 1} \paren {x - 1} }\)

\(\ds \)

\(=\)

\(\ds \frac {x - 3} {\paren {x + 1} \paren {x - 1} }\)

We can see that $\map {g'} x \ge 0$ for $x \ge 3$.

So from Real Function with Positive Derivative is Increasing:

$g$ is increasing.

So for $x \ge 3$ we have:

\(\ds \paren {x - 1} \map \ln {x - 1} - x \ln x + 2 \map \ln {x + 1}\)

\(\ge\)

\(\ds \map g 3\)

\(\ds \)

\(=\)

\(\ds 2 \ln 2 - 3 \ln 3 + 2 \ln 4\)

\(\ds \)

\(=\)

\(\ds 6 \ln 2 - 3 \ln 3\)

Logarithm of Power

\(\ds \)

\(=\)

\(\ds \map \ln {\frac {64} {27} }\)

Logarithm of Power

so for $x \ge 3$ we have:

\(\ds \paren {x - 1} \map \ln {x - 1} - x \ln x\)

\(\ge\)

\(\ds -2 \map \ln {x + 1} + \map \ln {\frac {64} {27} }\)

\(\ds \)

\(>\)

\(\ds -2 \map \ln {x + 1}\)

$\blacksquare$