Inequality between Neumann and Dirichlet eigenvalues Let $\Omega$ be a fixed, smooth and bounded domain in $\mathbb{R}^n$. If we denote with $\{\lambda_n\}_{n \ge 1}$ the nondecreasing sequence of eigenvalues of the Dirichlet problem 
$$\left\{
\begin{array}{rc} -\Delta u = \lambda u & \Omega \\ u=0 & \partial\Omega \end{array}\right.$$
and we denote with $\{\mu_n\}_{n \ge 1}$ the nondecreasing sequence of eigenvalues of the Neumann problem
$$\left\{\begin{array}{rc} -\Delta u = \mu u & \Omega \\ \frac{\partial u}{\partial \nu}=0 & \partial\Omega \end{array}\right.$$
then we have 
$$\tag{1}\mu_j\le \lambda_j,\quad \forall j=1, 2, \ldots$$
I would like to prove (1). 
The page at the Encyclopaedia of Mathematics invokes the max-min characterization: both for Dirichlet and for Neumann eigenvalues we have 
$$\lambda_k=\sup \inf \frac{\int \lvert \nabla u\rvert^2}{\int u^2} [=\mu_k],$$
where the $\inf$ is taken over all $u\in H^1_0$ [resp. $H^1$] which are orthogonal to $\phi_1\ldots \phi_{k-1}$ and the $\sup$ is taken over all possible choices of $\phi_1\ldots \phi_{k-1}\in H^1_0$ [resp. $H^1$]. 

Question. According to the linked page, inequality (1) follows "obviously" from this characterization. However, this does not seem that obvious to me, as the $\sup$ for the Neumann eigenvalues is taken over a larger class. Can someone explain me how is this done?

Thank you for reading.
 A: It has to do with the way the different boundary value problems are formulated. In one, you are considering functions lying in $H^1_0$, in the other, over functions lying in $H^1$. Recall that $H^1_0$ is a subset of $H^1$. 
(edited after commentary):
Aha, I realized. My answer is still correct, but the formulation needs to be revised. The sup is taken over all $\phi$ in $H^1$ in both cases, in which case it's only the conditions on the inf part that change between the Dirichlet and Neumann eigenvalues, which lets you compare apples to apples, so to speak. 
Taking 
$$\lambda_k = \sup \inf \frac{\int |\nabla u|^2}{\int u^2}$$
where we take $u\in H^1_0$ orthogonal to $\phi_1 \ldots \phi_{k-1}$ lying inside $H^1$ still yields the traditional Dirichlet eigenvalue. It is clear that when the $\phi_i$ are the $k-1$ Dirichlet eigenfunctions, $\lambda_k$ is achieved by the ratio
$$D[\phi_1,\ldots,\phi_{k-1}] = \inf \frac{\int |\nabla u|^2}{\int u^2}$$
We need only show that for any other choice of the $\phi_i$, the ratio achieved is $\leq \lambda_k$. Morally, the argument goes like this: suppose the space of functions spanned by the $\phi_i$ does not include the eigenspace corresponding to $\lambda_1$; then the natural value for the ratio is $\lambda_1 \leq \lambda_k$. You repeat inductively on each of the eigenspaces corresponding to eigenvalues, until you have specified $k-1$ functions, and then what's left must be $\lambda_k$. 
