I am looking for a graph for which $2 d\_{i} < \mu\_{i}$, for some index $i$, where $\mu\_{1} \leq \mu\_{2} \leq \dots\leq \mu\_{n}$ are the eigenvalues of the Laplacian matrix $L(G)$ and $d\_{1} \leq d\_{2} \leq \dots \leq d\_{n}$ are the node degrees. 


According to the literature and existing upper/lower bounds on the eigenvalues of the Laplacian as a function of node degrees, it seems 
there is a graph with this property. However, I was not able to find such a graph by generating all graph with $4, 5, \dots, 10$ vertices. 
Any help or advice for a possible hypothesis confirmation or rejection would be appreciated.


Related references:


* Miriam Farber, Ido Kaminer, [Upper bound for the Laplacian eigenvalues of a graph](https://arxiv.org/abs/1106.0769), June 2011.
* A. E. Brouwer, W. H. Haemers, [A lower bound for the Laplacian eigenvalues of a graph—proof of a conjecture by Guo](https://pdfs.semanticscholar.org/f10a/e559b9ff043ef0492df8f15722b3d16a1bd8.pdf), February 2008.



Lower bound on the eigenvalues of the Laplacian
This is an expanded solution based on the comment of @mostafa.


Such graphs dose not exist.


Let $L$ be the Laplacian of the graph. Suppose that diagonal elements of $L$ are sorted
sequence of degrees $d\_1\leq \ldots \leq d\_n$. 
 By [Min-Max theorem](https://en.wikipedia.org/wiki/Min-max_theorem#Min-max_theorem), we have


$$\mu\_i= \min\_{U<\mathbb R^n,\dim(U)=i} \big\{ \max\_{x \in U, \|x\|=1} x^TL x \big\} $$


So, if there exists an $i$-dimensional subspace of $\mathbb R^n$, say $U$, such that


$$
\max\_{x \in U, \|x\|=1} x^TL x \leq 2d\_i
$$
then we have $\mu\_i \leq 2d\_i$. 


Let $U := \langle e\_1,\ldots,e\_i\rangle$, where $e\_k$ is $k$-th element of the standard base of $\mathbb R^n$, for $k=1,\ldots,i$. Suppose that $x\in U$ and $\|x\|=1$. Note that only first $i$ elements of $x$ are nonzero. we have
$$
x^TL x = \sum\_{ab \in E(G),a<b} (x\_a-x\_b)^2 \leq \sum\_{ab \in E(G),a<b} 2(x\_a^2+x\_b^2)=2\sum\_{k=1}^id\_k x\_k^2\leq2d\_i.
$$
Now the statment at hand.


