I have two questions regarding the proof of Theorem 5.5. of Titchmarsh's book The Theory of the Riemann Zeta-Function:
Green-underlined: How from the $O$-term $$\mathcal{O} {\Bigg(\mu + \sum_{r=1}^{\mu -1} \min (\mu, |\csc \dfrac{tr}{2(a+ \nu \mu)^2}|\Bigg)}^{\frac12},$$ the green-underlined equality is achieved? I worked on this for a while and even my guess wouldn't help to achieve a rigorous proof: First by $(a+b)^{\frac12} \le a^{\frac12} + b^{\frac12}$ it brings the sqrt root inside (?) then instead of having $\sum_{\nu}^{N+1}$ behind the sqrt root it is inside it by comes twice (like $a = (a^2)^{\frac12}$ but is this reasonable?!
Orange-underlined: I couldn't understand the whole paragraph starting from "Now" which is underlined by an orange line. For more clearance, I explain my thoughts: A) I am not a native speaker; what does it mean "the least value but two of"? And why one of the sine terms is greater than $Ar / {\mu}^2$? B) The difference of two consecutive terms (i.e. the equality comes just after "Now") is roughly $t r \mu / a^3$ and it is roughly equals to $r / {\mu}^2$ so $tr / 2(a+ \nu \mu)^2$ are distributed on a line partitioned with intervals of length $r / {\mu}^2$, and what it has to do with $\sin$ of theses values?
Update: I have the answer for the first question: It's Holder's Inequality. Still the second question remains! My second question is asking for a simple clear explanation of what's going on the mentioned paragraph.