Here is a simple proof. Without loss of generality, $G'$ is obtained from $G$
by deleting some edges (and keeping all vertices). Let $A$ and $A'$ denote
the adjacency matrices of $G$ and $G'$, respectively, and let $x'$ be an
eigenvector of $A'$, belonging to the eigenvalue $\lambda_1(G')$, such that
all coordinates of $x'$ are non-negative. We have then
$$ \lambda_1(G') = \frac{\langle x',A'x'\rangle}{\|x'\|^2}
\le \frac{\langle x',Ax'\rangle}{\|x'\|^2}
\le \sup_{x\ne 0} \frac{\langle x,Ax\rangle}{\|x\|^2}
= \lambda_1(G); $$
indeed, if all coordinates of $x'$ are strictly positive, then the first
inequality is strict, and if $x'$ has zero coordinates, then the second
inequality is strict (by Perron-Frobenius, which says that the supremum is
attained on a vector with all coordinates distinct from $0$).

The largest *Laplacian* eigenvalue (which, of course, is equal to the Laplacian spectral radius) can be dealt with in a similar manner. Suppose that $G'$ is obtained from $G$ by deleting some edges, and let $E$ and $E'$ be the edge sets of $G$ and $G'$, respectively. Furthermore, denote by $\mu_n(G)$ and $\mu_n(G')$ the largest Laplacian eigenvalues of $G$ and $G'$, and fix an eigenvector $x'$ of $G'$, belonging to the eigenvalue $\mu_n(G')$. Indexing coordinates by the vertices of $G$, we have

$$ \mu_n(G') = \frac{\sum_{(u,v)\in E'} (x'_u-x'_v)^2}{\|x'\|^2}
\le \frac{\sum_{(u,v)\in E} (x'_u-x'_v)^2}{\|x'\|^2}
\le \sup_{x\ne 0} \frac{\sum_{(u,v)\in E} (x_u-x_v)^2}{\|x\|^2}
= \mu_n(G). $$

Notice that the inequality may fail to be strict in the Laplacian case; say, $\mu_3(K_3)=\mu_3(P_3)=3$.

An Introduction to the Theory of Graph Spectra. $\endgroup$ – David Bevan Oct 20 '14 at 12:31