Good morning everyone! Today I have a question that might appear trivial to you but since I am from non-math background, please bear with me.
The issue is, inequality involved monotonically increasing function. Let me clarify.
In the proof on p. 461 ff of the Sixth Edition of Hardy and Wright's An Introduction to the Theory of Numbers (Theorem 423) the following line is written directly -
$\int_{n-1}^{n}\log^h{(\frac x t)}dt \geq \log^h{(\frac x n)}$
after indicating that the function is monotonically increasing (see this post also) -
Since $\log t$ increases with $t$, we have, for $n > 2$,
Also in a very good answer(click here to see) written (involved with inequality) by very helpful Markus Scheuer, it is mentioned that -
The Von Mangoldt function Λ(n) takes non-negative values only, so that \begin{align*} \sum_{mn\leq x}\Lambda(m)\Lambda(n)\tag{$\ast\ast$} \end{align*} is monotonically increasing with x. and then implies inequality.
Now this is quite intuitive, and simple observation (also I have case specific proof, see this answer (click here), but not the general case), but in mathematics, everything is proved rigorously, then why it is mentioned casually in book without proof, and people are using that as a established fact?
It could be the case that it is proved in some book I am not aware of, in that case please mention, which book contains such proof of regarding the inequality of the monotonic function?
Thank you.
Generally, for an continuous function $f(x)$ over some interval $[a,b]$, we always have
$$ \min_{a\le x\le b}\{f(x)\}(b-a)\le\int_a^bf(x)\mathrm dx\le\max_{a\le x\le b}\{f(x)\}(b-a) $$
Because $\log^h\left(\frac xt\right)$ increases with $t$, its minimum value on interval $[n,n+1]$ occurs at $t=n$, thus
$$ \int_n^{n+1}\log^h\left(\frac xt\right)\mathrm dt\ge\log^h\left(\frac xn\right) $$
For the answer you've mentioned, the non-negativity of the summands suffices to show the increasing property of a partial summation. That is, if $a_n$ is non-negative, then
$$ \sum_{k=1}^Na_k $$
is monotonically increasing. I hope this will help you!