Let $n\geq 1$, $\epsilon>0$, $v\in \mathbb{R}^n$ and $c>0$. I want to prove that: $$\sum_{k=1}^n \left|v^2_{k+1}-v^2_k \right|\sqrt{\epsilon \frac{k}{n^{\frac{1}{3}}}+c} \leq \sqrt{\epsilon}n^\frac{1}{3}\sum_{k=1}^n (v_{k+1}-v_k)^2 + \frac{1}{\sqrt{\epsilon} n^{\frac{1}{3}}}\sum_{k=1}^n \left(\epsilon \frac{k}{n^\frac{1}{3}}+c\right)v^2_k .$$
This is from a textbook I am currently studying. I understand that we get the v's on the RHS thanks to $v_{k+1}^2-v_{k}^2 \leq (v_{k+1}-v_{k})^2 + 2v_{k+1}v_{k} $ and Cauchy-Schwarz (applied on $v_{k+1}v_{k}$) but I am struggling for the whole bound.