Suppose $0=\lambda_0\le\lambda_1\le \ldots \le\lambda_n$ be the eigen values of the normalized laplacian of a graph $G$.
Show that $\lambda_1\ge \dfrac{1}{D\text{vol}G}$
where $D$ denotes the diameter of $G $ and $\text{vol}=$ sun of all vertex degrees.
How should I approach this?
Recall that:
$$\lambda_{1} = \inf_{f \perp T1} \dfrac{\sum_{u \sim v} (f(u) - f(v))^{2}}{\sum_{v \in V(G)} f(v)^{2} d_{v}}$$
This is just the Rayleigh quotient definition of $\lambda_{1}$. Now let $f$ be a Harmonic eigenfuction of $G$ achieving $\lambda_{1}$, and let $v_{0}$ be a vertex that is a solution to $\max_{v \in V(G)} |f(v)|$. As $f$ is a Harmonic eigenfunction, it is perpendicular to the degree vector. So there exists a vertex $u_{0}$ such that $f(v_{0})f(u_{0}) < 0$. Let $P$ be a shortest $v_{0}-u_{0}$ path in $G$. We have:
$$\lambda_{1} = \inf_{f \perp T1} \dfrac{\sum_{u \sim v} (f(u) - f(v))^{2}}{\sum_{v \in V(G)} f(v)^{2} d_{v}} \geq$$
$$\dfrac{\sum_{xy \in P} (f(x) - f(y))^{2} }{\text{vol}(G) f(v_{0})^{2}}$$
This follows from the fact that we have this inequality (as $P$ has no more edges than $G$): $$\sum_{u \sim v} (f(u) - f(v))^{2} \geq \sum_{xy \in P} (f(x) - f(y))^{2}$$
And as $v_{0}$ is a maximizer for $|f|$, we have:
$$\sum_{v \in V(G)} f(v)^{2} d_{v} \leq \text{vol}(G) f(v_{0})^{2}$$
We now apply:
$$\sum_{xy \in P} (f(x) - f(y))^{2} \geq (f(v_{0}) - f(u_{0}))^{2} \geq \frac{1}{D}(f(v_{0}) - f(u_{0}))^{2}$$
To obtain:
$$\dfrac{\sum_{xy \in P} (f(x) - f(y))^{2} }{\text{vol}(G) f(v_{0})^{2}} \geq$$
$$\dfrac{(f(v_{0}) - f(u_{0}))^{2}}{D \text{vol}(G) f^{2}(v_{0})}$$
As $f(v_{0})f(u_{0}) < 0$, $f(v_{0})^{2} - 2f(v_{0})f(u_{0}) + f(u_{0})^{2} > 0$. So:
$$\dfrac{(f(v_{0}) - f(u_{0}))^{2}}{D \text{vol}(G) f^{2}(v_{0})} \geq \frac{1}{D \text{vol}(G)}$$
Note that this proof is in Fan Chung's text on Spectral Graph Theory, the first four chapters of which are free online (http://www.math.ucsd.edu/~fan/research/cb/ch1.pdf). This proof is in Ch.1, but it takes some work to fill in the details.