Suppose that $1 \leq q,r\leq \infty$ and that $\frac{2}{p} = \frac{1}{r} + \frac{1}{q}$. Let $I$ be an interval. How can we show that
$$\int_I|u_x|^pdx \leq C^p|I|^{1+p-\frac{p}{r}}\left(\int_I|u_{xx}|^rdx \right)^{\frac{p}{r}} + C^p |I|^{-\left(1+p-\frac{p}{r} \right)}\left( \int_I u^qdx\right)^{\frac{p}{q}} $$
Could we assume $u_x\ge 0$ and $u=0$ on the boundary of I?
If so:
First Integrate by parts: $$ \int_I u_x^p \,dx = \int_I u_x^{p-1} \,u_x \,dx = (p-1) \int_I u\, u_x^{p-2}\,u_{xx} \, dx \le (p-1)\, \int_I | u|\, |\,u_x^{p-2}|\,|\,u_{xx}| \, dx =I $$
Now use Holder inequality (note that $\frac{1}{r}+\frac{1}{q}+(1-\frac{2}{p})=1$)
$$ \int_I u_x^p \,dx \le I\le (p-1) \left(\int_I |u|^q \right)^{1/q}\left(\int_I |u_{xx}|^r \right)^{1/r}\left(\int_I |u_x|^p \right)^{1-2/p} $$ which implies $$ \int_I |u_x|^p \le (p-1)^{p/2} \left(\int_I |u|^q \right)^{p/2q}\left(\int_I |u_{xx}|^r \right)^{p/2r} $$
Finally, use Cauchy inequality with $\varepsilon$.