The following is a Lemma on a paper I am reading (Existence of Solutions to a Singular Elliptic Equation - M. Montenegro - 2011). I am having some trouble with the conclusion of the proof and would be the most glad for any help.
First, a few definitions. Let $\Omega \subset \mathbb R^N$, $0 < q < p < 1$ and $0 < \beta < 1$ and define $$ g_\varepsilon(u) = \begin{cases} \frac{u^q}{(u + \varepsilon)^{q + \beta}}, \quad u \geq 0 \\ 0, \quad u < 0 \end{cases} $$ $$ G_\varepsilon(u) = \int_0^ug_\varepsilon(s) \ ds \geq 0 $$ and $I_\varepsilon: H_0^1(\Omega) \to \mathbb R$ by $$ I_\varepsilon(u) = \frac 12 \int_\Omega |\nabla u|^2 \ dx + \int_\Omega G_\varepsilon(u) \ dx - \frac{\lambda}{p + 1} \int_\Omega(u^+)^{p + 1} $$
Lemma: Given $\lambda > 0$, there exist $a_2, b_2 > 0$ and $0 < \rho < 1$ such that for every $0 < \varepsilon < 1$ $$ I_\varepsilon(u) \geq a_2 > 0 \quad \forall u \text{ such that } \|u\|_{H_0^1} = \rho, $$ the functional $I_\varepsilon$ is coercive and bounded from below by $-b_2$.
Proof:[As in the paper] From the definition of $g_\varepsilon$, we have that $$ g_\varepsilon (t) \geq \frac{t^q}{(t + 1)^{q + \beta}} $$ for every $t \geq 0$. Since $q < p$ we may find $\delta(\lambda)$ such that $$ g_\varepsilon(t) \geq \lambda t^p \quad \forall 0 \leq t \leq \delta. $$ (for a proof, see this question) Taking $\rho$ sufficiently small, we obtain $I_\varepsilon(u) \geq a_2 := \rho^2/4$ for every $u$ such that $\|u\|_{H_0^1} = \rho$. We also obtain that $$ I_\varepsilon(u) \geq \frac12 \|u\|_{H_0^1}^2 - C\|u\|_{H_0^1}^p, $$ from which the boundedness and coercivity follow.
My questions:
How can we conclude that $I_\varepsilon(u) \geq \rho^2/4$ for all $\|u\|_{H_0^1} = \rho$? I have no clue on that.
Does the last inequality follow because $\|u^+\|_p^p \leq \|u\|_p^p \leq \|u\|_{H_0^1}^p$?
Thanks in advance and kind regards.
I found an answer to both questions in this more detailed paper by the same author together with Elves Silva: Two solutions for a singular elliptic equation by variational methods