A consequence of the Min-Max Principle for self-adjoint operators Let $H=(H, (\cdot, \cdot))$ be a Hilbert space. Let $T_1,T_2:D \subset H \longrightarrow H$ be a self-adjoint operators (not necessarily bounded). It's well-know that the spectrum $\sigma(T_i)$ of $T_i$ satisfies $\sigma(T_i) \subset \mathbb{R}$, for $i=1,2$ (see Theorem $29.2$ in $[3]$). Suppose that $T_1$ and $T_2$ are bounded below and has $N \in \mathbb{N}$ (real) eigenvalues arranged in the ascending order
$$
\lambda_1(T_i) \leq \lambda_2(T_i) \leq \lambda_3(T_i) \leq \cdots \lambda_N(T_i), \quad i \in \{1,2\}.
$$
As a consequence of the Min-Max Principle $($see $[2$, page $85]$ or $[1$, page $61])$, if
$$
(T_1(u), u) \leq (T_2(u), u),\; \forall \; u \in D \tag{1}
$$
then,  for each $n \in \{1,\cdots, N\}$,
$$\lambda_n(T_1) \leq \lambda_n(T_2). \tag{2}$$
Question. If
$$
(T_1(u), u) < (T_2(u), u),\; \forall \; u \in D\setminus \{0\}
$$
and
then
$$\lambda_n(T_1) < \lambda_n(T_2) \tag{3}
$$ for each $n \in \{1,\cdots, N\}$?
I think so, because the Min-Max Principle establishes that, for $i=1,2$,
$$
\lambda_n(T_i)= \sup_{u_1, u_2, \cdots u_{n-1} \in H } \inf_{v \in D\setminus \{0\} \atop v \in [u_1, u_2, \cdots u_{n-1}]^{\perp} } \frac{(T_i(v),v)}{\|v\|}.
$$
Remark. I did this question in Math Stackexchange, but I don't received any comment or answer.
Any comment or reference are welcome.
$[1]$  Kato, T., Perturbation Theory for Linear Operators, $2$nd edition, Springer, Berlin, $1984$.
$[2]$ Reed, S. and Simon, B., Methods of Modern Mathematical Physics: Analysis of Operator,
Academic Press, Vol. IV, $1978$.
$[3]$ Bachman, G. and Narici, L. Functional Analysis. New York: Academic Press, $1966$.
 A: I'm expanding my comment, in response to the OP's comment. Indeed, the case of just the lowest eigenvalue is perhaps not a good illustration of the full argument.
In general, let $u_j$ be a normalized eigenvector for $\lambda_j(T_1)$, so $T_1 u_j=\lambda_j(T_1) u_j$. Also make sure that the $u_j$ are orthogonal (this is automatic, except in the case of degeneracies). Then
$$
\lambda_n(T_1)=\langle u_n, T_1 u_n\rangle =\inf_{v\perp u_1,\ldots , u_{n-1}} \langle v, T_1 v\rangle < \inf _{v\perp u_1,\ldots , u_{n-1}} \langle v, T_2 v\rangle .
$$
The inequality is true because the infima are really minima: by the assumption on the existence of discrete spectrum in the range we're investigating and by the spectral theorem, the search for $v$ can be restricted to a suitable finite-dimensional subspace. We can then try the $v$ that minimizes $\langle v, T_2 v\rangle$ in the other quadratic form.
We're done since obviously $\lambda_n(T_2)\ge \inf _{v\perp u_1,\ldots , u_{n-1}} \langle v, T_2 v\rangle$.

The argument can also be organized differently, maybe this version is more transparent: Let $M\subseteq H$, $\dim M=n$, be the ("an", in the case of degeneracies) space spanned by the eigenvectors of $T_2$ with eigenvalues $\lambda_1(T_2),\ldots , \lambda_n(T_2)$. Then $\max_{v\in M, \|v\|=1 }\langle v, T_2 v\rangle = \lambda_n(T_2)$.
By assumption and since $M$ is finite-dimensional,
$$
t=\max_{v\in M, \|v\|=1 } \langle v, T_1 v\rangle < \lambda_n(T_2) .
$$
For any choice of $u_1,\ldots , u_{n-1}\in H$, there will be a $v\in M\ominus H$, $\|v\|=1$. Hence, by min-max, $\lambda_n(T_1)\le t<\lambda_n(T_2)$.
