For $\varOmega \subset \mathbb{R}^n$, $n\ge2$, a bounded domain and $1 < q < 2 < p < \frac{2n}{(n-2)_+}$ I want to show that \begin{equation*} \| u \|_{L^p(\varOmega)}^2 \le C \left( \| \nabla u \|_{L^2(\varOmega)}^2 + \| u \|_{L^q(\varOmega)}^2 \right) \end{equation*} for all $u \in W^{1,2}(\varOmega)$ and some $C > 0$.
I have $\frac1p > \frac{n-2}{2n} = \frac12 - \frac1n$ and from Sobolev embedding \begin{equation*} \| u \|_{L^{p}(\varOmega)}^2 \le C_1 \left( \| \nabla u \|_{L^2(\varOmega)}^2 + \| u \|_{L^2(\varOmega)}^2 \right). \end{equation*}
I don't know how to conclude the above inequality from this?
By Hölder's inequality, with $\theta = \frac{1/q-1/2}{1/q-1/p} ∈ (0,1)$, it holds $$ \|u\|_{L^2} ≤ \|u\|_{L^p}^θ \|u\|_{L^q}^{1-\theta} =: a^θ b^{1-\theta} $$ and so, since by Young's inequality, for any $\varepsilon>0$, $$ a^{2θ} b^{2(1-\theta)} = (\varepsilon a^2)^θ \, \varepsilon^{-\theta}\, b^{2(1-\theta)} ≤ \varepsilon \, \theta\, a^2 + \frac{(1-\theta)}{\varepsilon^{\theta/(1-\theta)}}\, b^2. $$ Starting from your second inequality, one obtains \begin{align*} \|u\|_{L^p}^2 &\leq C_1 \left(\|\nabla u\|_{L^2}^2 + \|u\|_{L^2}^2 \right) \\ &\leq C_1 \|\nabla u\|_{L^2}^2 + \varepsilon \, \theta\,C_1\, \|u\|_{L^p}^2 + \frac{(1-\theta)\,C_1}{\varepsilon^{\theta/(1-\theta)}}\, \|u\|_{L^q}^2 \end{align*} and so removing $\varepsilon \, \theta\,C_1\, \|u\|_{L^p}^2$ from both sides leads to \begin{align*} (1-\varepsilon \, \theta\,C_1)\,\|u\|_{L^p}^2 &\leq C_1 \|\nabla u\|_{L^2}^2 + \frac{(1-\theta)\,C_1}{\varepsilon^{\theta/(1-\theta)}}\, \|u\|_{L^q}^2. \end{align*} Taking any $ε < 1/(θ\, C_1)$ and dividing by $1-\varepsilon \, \theta\,C_1 > 0$ on both sides finally yields \begin{align*} \|u\|_{L^p}^2 &\leq C \left( \|\nabla u\|_{L^2}^2 + \|u\|_{L^q}^2\right), \end{align*} with $C = \frac{C_1}{1-\varepsilon \, \theta\,C_1}\max\!\left(1,\frac{(1-\theta)}{\varepsilon^{\theta/(1-\theta)}}\right)$.