The motivation of the question is the Chebyshev Bound in the context of Analytic Number Theory but I do not think the context matters too much here, nevertheless, the statement was $\exists C> 0$ such that for all $x\geq 2$, $\pi(x)\leq C\frac{x}{log x}$.
Question: I was wondering how to prove that there exists a $C$, positive, such that $\forall x\geq 2$ this inequality holds: $$\frac{\frac{x}{2}}{log(\frac{x}{2})}+\frac{\frac{x}{4}}{log(\frac{x}{4})}+\frac{\frac{x}{8}}{log(\frac{x}{8})}+\cdots\leq C\frac{x}{log(x)}?$$
I did some digging on the internet and I found this quite useful link and I tried to follow Exercise 8. I think I managed to do it but here is my issue: it felt like that I was doing '$\forall x, \exists C>0$ such that the above inequality holds' rather than '$\exists C>0$ such that $\forall x$ the inequality holds'.
So if someone could please help me with proving this inequality, it would be much appreciated as the method I linked above did not convince enough.
Thank you so much in advance!