Inequality $\int^{1}_{0}(u(x))^2\,\mathrm{d}x \leq \frac{1}{6}\int^{1}_{0} (u'(x))^2\,\mathrm{d}x+\left(\int_{0}^{1} u(x)\,\mathrm{d}x \right)^2$ I've been scratching my brain on this one for about a week now, and still don't really have a clue how to approach it.
Show that for $u \in C^1[0, 1]$ the following inequality is valid:
$$\int^{1}_{0}(u(x))^2\,\mathrm{d}x \leq \frac{1}{6}\int^{1}_{0} (u'(x))^2\,\mathrm{d}x+\left(\int_{0}^{1} u(x)\,\mathrm{d}x \right)^2$$

The constant $\frac{1}{6}$ can be slightly improved without using the full strength of Wirtinger Inequality.
$$\begin{align} \int_0^1 u^2(t)\,dt - \left(\int_0^1 u(t)\,dt\right)^2 & = \frac{1}{2}\int_0^1\int_0^1 \left(u(x) - u(y)\right)^2\,dx\,dy \\ &= \int_0^1 \int_x^1 \left(\int_x^y u'(t)\,dt\right)^2 \,dy\,dx \\ & \le \int_0^1\int_x^1 (y-x)\left(\int_x^y (u'(t))^2\,dt\right) \,dy\,dx \tag{1} \\ & = \int_0^1 (u'(t))^2 \int_0^t \int_t^1 (y-x)\,dy\,dx\,dt \tag{2} \\ & = \frac{1}{2}\int_0^1 t(1-t)(u'(t)) ^2 \,dt \\ & \le \frac{1}{8} \int_0^1 (u'(t))^2\,dt \tag{3}\end{align}$$
$(1)$ Cauchy Schwarz Inequality.
$(2)$ Change in order of integration.
$(3)$ $\displaystyle t(1-t) \le \frac{1}{4} \textrm{ for } t \in (0,1)$.
