I am trying to show that if $\Omega \subset\subset \mathbb{R}^2$ with $C^1$ boundary and $ u \in W^{1,2}(\Omega)$ then
\begin{equation*} \int u^4 < C \left(\int u^2 \right)^2 + C \left(\int u^2 \right) \left(\int |Du^2|\right) \end{equation*}
for some constant $C$ depending on $\Omega$, where all integrals are over $\Omega$.
I feel as though I am almost there. As $u \in W^{1,2}$ we have that $u^2 \in W^{1,1}$ so by the Sobolev embedding theorem
\begin{align*} ||u^2||_{L^2} &\leq C||u^2||_{W^{1,1}} \\ &=C||u^2||_{L^1} + C||D(u^2)||_{L^1} \\ &=C||u^2||_{L^1} + C||(u \cdot Du)||_{L^1} \\ &=C||u^2||_{L^1} + C||u||_{L^2}||Du||_{L^2} \end{align*}
where Holder's inequality has been used in the last line. This then reads
\begin{equation*} \left(\int u^4 \right)^{\frac{1}{2}} < C \left(\int u^2 \right) + C \left(\int u^2 \right)^{\frac{1}{2}} \left(\int |Du^2|\right)^{\frac{1}{2}} \end{equation*}
which when I square both sides gives nearly what I want but also introduces an unwanted cross term. Is there a way I can get around this? I don't think I can just increase the constants on the terms I want as then $C$ would depend on $u$.
Note that $(x+y)^2\le 2(x^2+y^2)$. More generally, if $p\in [1,\infty)$, we have that $$(x+y)^p\le 2^{p-1}(x^p+y^p), \ \forall\ x,y\ge 0.$$