Let $\Omega$ be a bounded Lipschitz domain. Can someone tell me what this inequality is called, or how to prove it:
$$\lVert u \rVert_{L^{r_1}(0,T;L^{q_1}(\Omega))} + \lVert u \rVert_{L^{r_2}(0,T;L^{q_2}(\Omega))} \leq C\left(\sup_{t \in [0,T]}\int_\Omega u(t)^2 + \int_0^T\int_\Omega |\nabla u|^2\right)^{\frac 12}$$
where $$\frac{1}{r_1} + \frac{N}{2q_1} = \frac{N}{4}$$ and $$\frac{1}{r_2} + \frac{N-1}{2q_2} = \frac{N}{4}.$$
I would think of this inequality as an interpolation inequality. In fact, you want to show $$\|u\|_{L^{r_1}(0,T;L^{q_1}(\Omega))} \le C \, \big(\|u\|_{L^\infty(0,T;L^2(\Omega))} + \|u\|_{L^2(0,T;H^1(\Omega))}\big).$$ By the Sobolev embedding theorem, it is sufficient to show $$\|u\|_{L^{r_1}(0,T;L^{q_1}(\Omega))} \le C \, \big(\|u\|_{L^\infty(0,T;L^2(\Omega))} + \|u\|_{L^2(0,T;L^{2\,N/(N-2)}(\Omega))}\big).$$
Using an interpolation argument, you get $$\|u\|_{L^{r_1}(0,T;L^{q_1}(\Omega))} \le \|u\|_{L^\infty(0,T;L^2(\Omega))}^{1-\lambda} \|u\|_{L^2(0,T;L^{2\,N/(N-2)}(\Omega))}^\lambda$$ for $\lambda \in [0,1]$ such that \begin{align*} \frac1{r_1} &= \frac{1-\lambda}{\infty} + \frac{\lambda}{2} \\ \frac1{q_1} &= \frac{1-\lambda}{2} + \frac{\lambda \,(N-2)}{2\,N} \end{align*} and Young's inequality yields the above inequality.
The same can be done with $r_2$ and $q_2$ (note that you can drop the $-1$ in your last equation).