Arithmetic Average of Second Chebyshev Function/Lemma 1

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

Then:


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

Proof
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:

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

So, from Real Function with Negative Derivative is Decreasing:


 * $f$ is decreasing.

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:


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