The problem is
Suppose $\Omega\subset\mathbb{R}^{2}$ is open, $u\in H_{0}^{1}(\Omega)$. Proof that$$\lVert u\rVert_{L^{4}(\Omega)}^{4}\leq4\lVert u\rVert_{L^{2}(\Omega)}^{2}\lVert\nabla u\rVert_{L^{2}(\Omega)}^{2}.$$
What I have done:
Since $C_{c}^{\infty}(\Omega)$ is dense in $H_{0}^{1}(\Omega)$, assume first $u\in C_{c}^{\infty}(\Omega)$. Integrating by parts shows$$\int_{-\infty}^{x}\dfrac{\partial u}{\partial x}(t,y)u(t,y)\,dt=u^{2}(x,y)-\int_{0}^{x}\dfrac{\partial u}{\partial x}(t,y)u(t,y)\,dt.$$So$$\lvert u(x,y)\rvert^{2}=2\int_{0}^{x}\dfrac{\partial u}{\partial x}(t,y)u(t,y)\,dt\leq2\int_{\mathbb{R}}\left| \dfrac{\partial u}{\partial x}(t,y)\right| \cdot\lvert u(t,y)\rvert\,dt,$$ a similar discussion then shows$$\lvert u(x,y)\rvert^{4}\leq4\int_{\mathbb{R}}\left| \dfrac{\partial u}{\partial x}(t,y)\right| \cdot\lvert u(t,y)\rvert\,dt\cdot\int_{\mathbb{R}}\left| \dfrac{\partial u}{\partial y}(x,t)\right| \cdot\lvert u(x,t)\rvert\,dt.$$Now use Schwarz's inequality,$$\lvert u(x,y)\rvert^{4}\leq4\left( \int_{\mathbb{R}}\left| \dfrac{\partial u}{\partial x}(t,y)\right| ^{2}\,dt\right) ^{\frac{1}{2}}\left( \int_{\mathbb{R}}\lvert u(t,y)\rvert^{2}\,dt\right) ^{\frac{1}{2}}\left( \int_{\mathbb{R}}\left| \dfrac{\partial u}{\partial x}(x,t)\right| ^{2}\,dt\right) ^{\frac{1}{2}}\left( \int_{\mathbb{R}}\lvert u(x,t)\rvert^{2}\,dt\right) ^{\frac{1}{2}}.$$
Finally, integrating over $\Omega$ should give the conclusion, but I can't deal with the right hand side. Any advice is appreciated.