Given a k-regular graph $G$ (every vertex is of degree k), one defines its Laplace operator as
$L(G)=D-A=kI-A$, where $I$ is identity matrix and $A$ adjacency matrix of $G$.
Let $\lambda\_{1}\leq \ldots \leq \lambda\_{n}$ be eigenvalue of $L(G)$, 
are there any results on lower bound on $\lambda\_{n}$. Of course, here I have in mind 
bounds other than a trivial one: $k\frac{n}{n-1}\leq \lambda\_{n}$



Bounds on maximal eigenvalue of a k-regular graph
Even though the question has been answered, I feel I should flesh out my comment into an answer. Apologies for the redundancy.


First, here's a simple proof of this special case of the Hoffman bound. Suppose $X$ is an independent set of vertices of $G$ (some call this a coclique) of cardinality $\alpha$. Let $\mathbf{x} = (x\_1, \ldots, x\_n)$ be the characteristic function of $X$, so $x\_i = 1$ if the $i$th vertex is in $X$, and $x\_i = 0$ if it isn't in $X$. Put $\mathbf{y} = \mathbf{x} - \frac{\alpha}{n} \mathbf{1}$.


Direct computation shows that $\langle \mathbf{y}, \mathbf{y} \rangle = \alpha - \frac{\alpha^2}{n}$. A slightly more bothersome calculation (using the fact that $\langle A \mathbf{x}, \mathbf{x} \rangle = 0$) shows that 
$$
\langle L\mathbf{y}, \mathbf{y} \rangle = k\langle \mathbf{y}, \mathbf{y} \rangle - \langle A\mathbf{y}, \mathbf{y} \rangle = k\left(\alpha - \frac{\alpha^2}{n}\right) - \left(0 - 2\frac{k \alpha^2}{n} + \frac{k\alpha^2}{n} \right) = k\alpha.
$$
Finally, since $\langle L \mathbf{y}, \mathbf{y} \rangle \leq \lambda\_n \langle \mathbf{y}, \mathbf{y} \rangle$, we see $\lambda\_n \geq k \frac{n}{n-\alpha}$ as desired.


Note in particular that since any graph which is regular of degree $k$ is $(k+1)$-colorable, there is an independent set of cardinality at least $\frac{n}{k+1}$. Plugging this in for $\alpha$ recovers the $\lambda\_n \geq k+1$ bound shown by Kevin in his answer.


