To prove that $$\pi(n)<6\frac{n}{\log n},$$ we examine the fact that $$4\log n>\log\binom{2n}{n}=\sum_{p\leqslant n}\log p\sum_{m=1}^{\left[\frac{\log n}{\log p}\right]}\left(\left[\frac{2n}{p^m}\right]-2\left[\frac{n}{p^m}\right]\right)\geqslant \sum_{n\leqslant p\leqslant 2n}\log p=\vartheta(2n)-\vartheta(n).$$ Then, we plugin $n=2^k$ to get $$\vartheta(2n)-\vartheta(n)<2^{k+1}\log 2$$ Summing on $k=0,1,2,\dots,r$, we have $$\vartheta(2^{k+1})<2^{k+2}\log 2$$ which can be further used to prove the desired inequality.
To prove that if $\sum_{n\leqslant x}a(n)\left[\frac{x}{n}\right]=x\log x+O(x)$, then $\sum_{n\leqslant x}a(n)\leqslant Bx$ where $B$ is a positive constant, we examine the difference $S(n)-S(n/2) $ where $S=\sum_{n\leqslant x}a(n)$. It is possible to prove that $$S(x/2^k)-S(x/2^{k+1})\leqslant K\frac{x}{2^k}$$ with the help of $T=\sum_{n\leqslant x}a(n)\left[\frac{x}{n}\right]$. Then, like what we did to $\vartheta$, sum on all $k=0,1,2,\dots$. Then we complete the proof.
These two instances show the use of the difference $f(2n)-f(n)$ in finding the bound of $f(n)$. Is there a specific name for this kind of technique? Where else can this technique be applied (other than finding upper and lower bound)? When should we consider to use this technique?
I don't know if these questions have answers, but I appreciate it if you can feed me some information about this technique. Thanks in advance. Any help will be appreciated.