Arithmetic Average of Second Chebyshev Function/Lemma 1

Lemma

Let $x \ge 1$ be a real number.

Then:

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

Define a function $f : \hointr 1 \infty \to \R$ by:

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

for each $x \in \hointr 1 \infty$.

Note that:

$\map f 1 = 2 \ln 2 - 1 \ln 1 - 2 \ln 2 = 0$

So it suffices to show that $f$ is decreasing for $x \ge 1$, then we will have $\map f x < 0$ for $x \ge 1$.

Note that $f$ is differentiable and:

$\ds \map {f'} x$

$=$

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

$\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 $

$\le$

$\ds \frac 1 x - \frac 2 {x + 1}$

$\ds $

$=$

$\ds \frac {1 - x} {x \paren {x + 1} }$

We can now see that $\map {f'} x \le 0$ for $x \ge 1$.

So, for $x \ge 1$, we have:

$\paren {x + 1} \map \ln {x + 1} - x \ln x - 2 \map \ln {x + 1} \le \map f 1 = 0$

That is:

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

$\blacksquare$