<s>[INST] Prove the inequality and limitation 
Suppose $\{a_k\}$ ,$\{b_k\}$ and $\{\xi_k\}$ are non negative, and for all $k\ge 0$ , we have $$a_{k+1}^2\le(a_k+b_k)^2-\xi_k^2$$
  Prove
  1.$$\sum_{i=1}^{k}\xi_i^2\le(a_1+\sum_{i=0}^kb_i)^2$$
  2. If $\{b_k\}$ also satisfy $\sum_{k=0}^{\infty}b_k^2\lt+\infty $, then $$\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^k\xi_i^2=0$$

After doing this transformation:$$\xi_k^2\le(a_k+b_k)^2-a_{k+1}^2$$
$\Rightarrow$ $$\sum_{i=1}^{k}\xi_k^2\le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+\cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$
then I can't move forward any more.
 [/INST] Let us first prove by induction that
$$\tag{1} a_1 +\sum_{i=0}^k b_i \ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 \le i \le k$, then we have
$$a_1 + \sum_{i=0}^{k+1} b_i \ge a_{k+1} + b_{k+1} \ge a_{k+1} + \big(\sqrt{a_{k+2}^2 + \xi_{k+1}^2} - a_{k+1} \big) \ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$\tag{2}\sum_{i=1}^k \xi_i^2 \leq \big(a_1 + \sum_{i=0}^k b_i \big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 \le (a_1+b_1)^2-\xi_1^2 \le (a_1+b_0+b_1)^2 - \xi_1^2.$$
Assume that (2) is true for $k$, then
\begin{align}
\big(a_1 + \sum_{i=0}^{k+1} b_i \big)^2-a_{k+2}^2 &\ge \sum_{i=1}^k \xi_i^2 + 2 \big( a_1 +\sum_{i=0}^k b_k \big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \\
&\ge \sum_{i=1}^k \xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \\
&\ge \sum_{i=1}^k \xi_i^2 + (a_{k+2}^2 +\xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \\
&\ge \sum_{i=1}^{k+1} \xi_i^2
\end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.
The previous result can be also applied with shifted-indices: The shifted sequence $(a_{n+j},b_{n+j},\xi_{n+j})_{n \in \mathbb{N}_0}$ satisfies the initial condition and therefore we can conclude (by using $(a+b)^2 \leq 2 a^2+2b^2$) that
\begin{align}
 \frac{1}{k-j+1} \sum_{i=j+1}^k \xi_i^2 &\le \frac{2}{k-j+1} \Big(a_{j+1}^2 + (\sum_{i=j}^k b_i)^2\Big)\\
&\le \frac{2a_{j+1}^2}{k-j+1} + 2\sum_{i=j}^k b_i^2,
\end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Assuming now additionally that $\sum_{n=1}^\infty b_n^2 <\infty$, we find that
\begin{align}
\limsup_{k \rightarrow \infty} \frac{1}{k}  \sum_{i=1}^k \xi_i^2 &\le \limsup_{k \rightarrow \infty} \frac{1}{k}  \sum_{i=1}^{j} \xi_i^2 + \limsup_{k \rightarrow \infty} \frac{1}{k}  \sum_{i=j+1}^k \xi_i^2 \\
&= \limsup_{k \rightarrow \infty} \frac{k-j+1}{k} \frac{1}{k-j+1}  \sum_{i=j+1}^k \xi_i^2 \\
&\le  \limsup_{k \rightarrow \infty} \frac{2a_{j+1}^2}{k-j+1} + 2\sum_{i=j}^k b_i^2 \\
&= 2\sum_{i=j}^\infty b_i^2 \rightarrow 0 \quad \text{for} \quad j \rightarrow \infty
\end{align}
proving the second part of the initial question.
</s>