An inequality with critical Sobolev exponent Let $\Omega\subset \mathbb{R}^n, n\geq 3$ be a nice bounded domain and $2^*=2n/(n-2)$ the critical Sobolev exponent. One may expect that $\forall \epsilon>0$, $\exists C_\epsilon<\infty$ such that 
$$
\|u\|_{2^*}\leq \epsilon \|\nabla u\|_2+C_\epsilon \|u\|_2, \quad \forall u\in H^1(\Omega).
$$
Due to the lack of  compact embedding from $H^1$ into $L^{2^*}$, the above inequality is indeed not true by the example listed this question.
Now, I wish to make it right by formulating it in a strengthened version as follows: given $p\in (2,2(n+2)/n)$ (or $p\in(2,2^*)$ in the worst case), $\forall \epsilon>0$, $\exists C_\epsilon<\infty$ such that 
$$
\|u\|_{2^*}\leq \epsilon (\|\nabla u\|_2+\|u\|_p^{p/2})+C_\epsilon(1+ \|u\|_2), \quad \forall u\in H^1(\Omega).   \tag{MCIS}
$$
I tried the example listed here, which does not give a counterexample. Also, arguing by contradiction seems not to work. Any help to prove or disprove (MCIS) is greatly acknowledged.
 A: Start with Pietro's example in the linked question. Fix $u$. Define 
$$ v_{\delta,M} = M\delta^{-n/2^*} u(x/\delta) $$
We know that $\|\nabla v_{\delta,M}\|_2 = M\|\nabla u\|_2$ and $\|v_{\delta,M}\|_{2^*} = M\|u\|_{2^*}$. Choose $\epsilon < \frac12 (\|u\|_{2^*} / \|\nabla u\|_2 )$. 
We also know that for $M$ fixed, as $\delta \to 0$ we have that $\|u\|_p \to 0$ in the range of $p$ you allow. 
Now, let $\Lambda > 0$ be arbitrary. We show that for every fixed $\Lambda$ there exists $\delta$ and $M$ such that 
$$ \|v_{\delta,M} \|_{2^*} > \epsilon (\|\nabla v_{\delta,M}\|_2 + \|v_{\delta,M}\|_p^{p/2} ) + \Lambda (1 + \|v_{\delta,M}\|_2) $$
which will disprove your desired inequality. 
By our choice of small $\epsilon$, it suffices to prove
$$ \frac12 \|v_{\delta,M} \|_{2^*} = \frac{M}{2} \|u\|_{2^*} > \epsilon \|v_{\delta,M}\|_{p}^{p/2} + \Lambda + \Lambda \|v_{\delta,M}\|_2 = \epsilon M \|v_{\delta,1}\|_p^{p/2} + \Lambda M \|v_{\delta,1}\|_2 + \Lambda $$
First choose $M$ sufficiently large that 
$$ \frac{M}{4} \|u\|_{2^*} > \Lambda $$
then choose $\delta$ sufficiently small so that the first two terms on the right hand side of the desired inequality is negligible. 
