This is question 8.23 part $4$ from H. Brezis Functional analysis
I already have that for any $f\in L^p(I)$, $p>1$ and $I=(0,1)$ there exists a unique $u\in H_0^1(I)$ satisfying $$\int_Iu'v'dx+k\int_Iuvdx=\int_Ifvdx$$ for all $v\in H_0^1(I)$, where $k>0$ is a constant.
Now I am trying to prove that $$\|u\|_{L^p}\leq \frac{1}{k+\delta/pp'}\|f\|_{L^p}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ where $\delta>0$ is independent of $k$ and $p$.
The hint suggests me to try $v=\gamma(u)$ where $\gamma(t):=|t|^{p-1}\text{sign}t$. I proceed with hint and I have the following result $$\int_I |u|^{p-2}(u')^2dx+k\int_I |u|^pdx\leq \left(\int_I |f|^p\right)^\frac{1}{p}\left(\int_I |u|^p\right)^{\frac{p-1}{p}} \,\,\,\,\,\,\,\,(2)$$
Then I try to compare $(2)$ and $(1)$, and I realize that I should try to prove the following inequality $$\int_I |u|^{p-2}(u')^2dx\geq C\int_I|u|^pdx\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$ and the constant $C$ should be very important. However, I don't have too much idea from here... Is it my direction is correct? What should I try next? Since the derivative of $u$ and $u$ is mixed in integration and all inequality I familiar with will give other side of inequality...
Update: It is easy to find that $$\|u\|_{L^p}\leq \frac{1}{k}\|f\|_{L^p}$$ and for trying to prove $(3)$, I wrote, by FTC, that $$\int_I |u|^pdx\leq p(p-1)\int_I\int_0^x\int_0^t |u(s)|^{p-2} |u'(s)|\,ds\,|u'(t)|\,dt\,dx$$ It looks very close to what I want... But again I have no idea what should I do next...
One idea would be to use $|u|^{p/2-1} u' = \frac p2(|u|^{p/2})'$. Then to estimate $$ \int_I |u|^{p-2} |u'|^2 dx= \frac{p^2}4 \int_I ((|u|^{p/2})')^2 \ge \frac{\pi^2 p^2}4\||u|^{p/2}\|_{L^2}^2 = \frac{\pi^2 p^2}4\|u\|_{L^p}^2. $$ Here, I used the from Poincare-Friedrichs inequality $$ \int_I (u')^2 \ge \pi^2 \int_I u^2 \quad u\in H^1_0(I). $$