Please help me to prove this statement:
If we have $0<\varepsilon<1$, $\frac{1}{m}<\frac{2}{2^{*}}$, where $2^{*}=\frac{2N}{N-2}$, with $N>2$ and $u_{\varepsilon}\geq 0$ satisfies \begin{equation} \begin{split} \left(\int_{\Omega}\left[(u_{\varepsilon}+\varepsilon)^{r}-1\right] \right)^{\frac{2}{2^{*}}}&\leq \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{\frac{1}{m}}, \end{split} \end{equation} then, $u_{\varepsilon}$ is uniformly bounded in $L^{r}(\Omega)$. Thank you in advance for your help.
Since \begin{equation} \left[\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}-1\right]^{\frac{2}{2^{*}}}\leq \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{\frac{1}{m}}, \end{equation} then, by using the fact that the real function $t\longmapsto t^{2^{*}}$ is increasing on $\mathbb{R}^{+}$, we have \begin{equation} \left[\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}-1\right]^{2}\leq \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{\frac{2^{*}}{m}}.\qquad(2) \end{equation}
Recalling the real inequality \begin{equation} \vert t^{2}-1\vert<\alpha(t-1)^{2}+\frac{1}{\alpha-1},\qquad\forall\alpha>1,\;\forall t\geq 0. \end{equation} Applying the above inequality for $t=\frac{1}{\vert\Omega\vert}\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}$ (this choice is possible since $u_{\varepsilon}\geq 0$ and $\varepsilon>0$), we get
\begin{equation} \begin{split} \left[\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}-1\right]^{2} &=\vert\Omega\vert^{2}\left[\left(\frac{1}{\vert\Omega\vert}\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)-1\right]^{2} \\ &>\frac{1}{\alpha}\big|\frac{1}{\vert\Omega\vert^{2}}\left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{2}-1\big|-\frac{1}{\alpha(1-\alpha)}.\qquad(4) \end{split} \end{equation} Combining (2) and (4), we deduce that \begin{equation} \big|\frac{1}{\vert\Omega\vert^{2}}\left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{2}-1\big|\leq\alpha \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{\frac{2^{*}}{m}}+\frac{1}{1-\alpha}.\qquad(5) \end{equation} (5) implies that \begin{equation} \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{2}\leq\alpha\vert\Omega\vert^{2} \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{\frac{2^{*}}{m}}+\frac{2-\alpha}{1-\alpha}\vert\Omega\vert^{2}.\qquad(6) \end{equation} Since $\frac{1}{m}<\frac{2}{2^{*}}$, then Young's inequality with $p=\frac{2m}{2^{*}}$ and $q=\frac{2m}{2m-2^{*}}$ on the first term on the right-hand side of (6) gives \begin{equation} \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{2}\leq \frac{2^{*}}{2m} \left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{2}+\frac{\alpha^{q}\vert\Omega\vert^{2q}}{q}+\frac{2-\alpha}{1-\alpha}\vert\Omega\vert^{2}.\qquad(7) \end{equation} From (7), we deduce that \begin{equation} \frac{2m-2^{*}}{2m}\left(\int_{\Omega}(u_{\varepsilon}+\varepsilon)^{r}\right)^{2}\leq \frac{\alpha^{q}\vert\Omega\vert^{2q}}{q}+\frac{2-\alpha}{1-\alpha}\vert\Omega\vert^{2}. \end{equation} So
\begin{equation} \Vert u_{\varepsilon}+\varepsilon\Vert_{r}\leq C, \end{equation} where C is positive constant independent of $\varepsilon$. Since $0<\varepsilon<1$ and the fact that \begin{equation} \big|\Vert u_{\varepsilon}\Vert_{r}-\Vert \varepsilon\Vert_{r}\big|\leq\Vert u_{\varepsilon}+\varepsilon\Vert_{r}, \end{equation} we deduce that \begin{equation} \Vert u_{\varepsilon}\Vert_{r}\leq C+\Vert\varepsilon\Vert_{r}\leq C+\vert\Omega\vert^{\frac{1}{r}}. \end{equation} So, the sequence $u_{\varepsilon}$ is uniformly bounded in $L^{r}(\Omega)$.