Let $X\subset\mathbb{R}^2$ be a bounded open set with either Dirichlet $u = 0$ or Neumann boundary $\frac{\partial u}{\partial \nu} = 0$ conditions on the boundary of $\Omega$. Let $V$ be either $H^1_0(X)$ or $H^1(X)$ (depending on the boundary condition) and take it as granted that the unique weak solution $u\in V$ to $-\Delta u + tu = f$ for some $f:X\to\mathbb{R}$ exists for $t > c$ for i.) in the case of Neumann c = 0, ii.) in the case of Dirichlet $c < 0$.
Given all this, if $\int_X\nabla u\cdot \nabla v = \lambda \int_X uv, v \in V$ has non-trivial solutions, why is it the case that $-\lambda \leq c$? This is what one lecture note I am reading states and I am not sure whether the inequality should be reversed, that $-\lambda \geq c$? The notes have not made any assumption on the sign of $t$ so in principle we could just take $t = -\lambda$ to get, by the claimed existence of the same notes, that as long as $-\lambda := t > c$, then such a unique $u\in V$ exists for $-\Delta u - \lambda u = f$.
The assumption is: $-\Delta u + tu = f$ is uniquely solvable for all $t > c$, $c$ given.
Now if there is $u\ne 0$ so that $\int_X \nabla u \nabla v = \lambda\int_X uv$ for all $v\in V$, then the equation $-\Delta u - \lambda u=0$ is not uniquely solvable, so that $-\lambda \le c$ follows