I am trying to solve the following question:
Let $Q$ be the cube $$\{0\leq x_i \leq \sigma\}\subset \mathbb{R}^n,$$ and $u\in C^{1}(Q)$. Then $$||u||_{L^2(Q)}^2\leq C\left[\frac{1}{\sigma^n}\left(\int_Q u(x) \ dx\right)^2+\sigma^2||u||^2_{H^1(Q)}\right]$$ where $C>0$ is a constant independent of $u$ and $\sigma$.
My attempt: I rewrite the above inequality. So it is equivalent to show that $$\int_Q|u|^2 \ dx\leq C\left[\left(\int_Q1\ dx\right)\left(\int_Q u(x) \ dx\right)^2+\sigma^2\left(\int_Q|u|^2 \ dx+\sum_i\int_Q\left|\frac{\partial u}{\partial x_i}\right|^2\ dx\right)\right]$$ Then I don't know how to continue... I want to apply Poincare inequality here (since the first term on the right side looks like the average of $u$ over $Q$) but it seems it does not work.
Let me do it for $n=1$, and leave the extension to higher dimensions to you, so you can have some fun. Let $$ A = \frac1\sigma\int_0^\sigma udx $$ be the average of $u$, and let $\xi\in[0,\sigma]$ be such that $u(\xi)=a$. Such $\xi$ exists because $u$ is continuous. Then we have $$ u(x)=A+\int_\xi^xu'(t)dt $$ which implies $$ |u(x)-A|\leq\int_0^\sigma|u'|\leq\sqrt\sigma\,\Big(\int_0^\sigma|u'|^2\Big)^\frac12 $$ or $$ |u(x)-A|^2\leq\sigma\int_0^\sigma|u'|^2 $$ by Cauchy-Bunyakowsky-Schwarz. Proceeding further, we get $$ |u(x)|^2\leq2A^2+2|u(x)-A|^2\leq2A^2+2\sigma\int_0^\sigma|u'|^2 $$ and integrating it over $x\in[0,\sigma]$ yields $$ \int_0^\sigma|u|^2\leq2\sigma A^2+2\sigma^2\int_0^\sigma|u'|^2 = \frac2\sigma\Big(\int_0^\sigma u\Big)^2 +2\sigma^2\int_0^\sigma|u'|^2. $$