Let $A\subset \{0,1,\dots, N-1\}$ such that $$|A\cap [0,N/3)|\geq \left(\delta+\dfrac{\delta}{8}\right)\cdot \dfrac{N}{3},$$ where $\delta\in (0,1]$. Prove that exists arithmetic progression $P$ with $|P|\ge N/3$ such that $|A\cap P|\geq \left(\delta+\dfrac{\delta}{8}\right)|P|$.

**Remark:** By $[0,N/3)$ I denote the set of integers from this interval.

If $N$ is divisible by $3$ then we can take $P:=[0,N/3)$ which is also a progression and $|P|=N/3$ we have $|A\cap P|\ge \left(\delta+\dfrac{\delta}{8}\right)|P|$.

If $N$ is not divisible by $3$ and if we we take $P:=[0,N/3)$ which is also progression and has $\lfloor\frac{N}{3}\rfloor+1$ elements. However, I am not able to derive the needed inequality.

Would be very grateful if anyone can explain how to approach in the case when $3\nmid N$?

P.S. This question refers to the proof of Roth's theorem.