Interpolation like inequality ,Question from Brezis' book exercise 8.15 I'm struggling with the following question from Haim Brezis' book:

Let $q\in[1,\infty)$ and $r\in (1,\infty]$. Show that there exists a constant $C=C(q,r)$ such that:
$\Vert u\Vert_{L^\infty(0,1)}  \leq C\cdot \Vert u \Vert_{W^{1,r}(0,1)}^a \cdot \Vert u \Vert_{L^{q}(0,1)}^{1-a} \quad \text{for all} \quad W^{1,r}(0,1)$
where $a\in (0,1)$ is determined by:
$a\Big( \frac{1}{q} +1- \frac{1}{r} \Big)= \frac{1}{q}$
And it gives the following hint:
When $u(0)=0$ write $G\big( u(x) \big)= \int_0^x G' \big( u(t) \big)u'(t)dt$ for $G(t)=\vert t \vert^{\alpha-1} \cdot t$ and $\alpha =\frac{1}{a}$. When $u(0)\neq 0$, use the previous case for $\zeta u$ where $\zeta \in C^1[0,1]$ satisfying $\zeta(0)=0$ and $\zeta(t)=1$ for $t\in [\frac{1}{2},1]$.

My attempts thus far: 
I know since $r>1$ that $u$ has a version which is in $C[0,1]$, and that if $r<\infty$ then it is also $(1-\frac{1}{r})$- Holder continuous. Also since $L^\infty(0,1)$ is embedded continuously into $C[0,1]$, there exists $C_1>0$ such that:
$ \Vert u \Vert_{L^\infty(0,1)} \leq C_1 \cdot \Vert u \Vert_{W^{1,r}(0,1)} \quad \text{for all} \quad u\in W^{1,r}(0,1) $
And because of this it now only remains to show that:
$\Vert u \Vert_{W^{1,r}(0,1)}^{1-a}\leq C_2 \cdot  \Vert u \Vert_{L^q(0,1)}^{(1-a)} \quad \text{for all} \quad u\in W^{1,r}(0,1) $
for some constant $C_2>0$. But I have not been able to understand where the hint comes into it at all. Also it seems to me that $G(t)$ is simply $t^\alpha$. I would appreciate any hints or help.
 A: I'll write my answer in case some one else will also want the solution for this question:
Let $u\in W^{1,r}(0,1)$. Every $v\in W^{(0,1)}$ satisfies that:
$v\in C[0,1]$ and $v(x)-v(y)=\int_y^x v'(t)dt$.
By a theorem stating that if $G(0)=0$ and $\vert G' \vert \leq M$ then:
$G\circ u \in W^{1,r}(0,1)$ and $\big( G\circ u \big)'(x)=G'\big( u(x)\big) \cdot u'(x)$, if $u(0)=0$ we can write therefore:
$G\circ u(x)= G\circ u(x)- G\circ u(0)= \int_0^x\big( G\circ u \big)'(t)dt$
Also $\vert G(u)\vert=\vert u\vert ^\alpha$, and thus:
$\vert u(x)\vert^\alpha= \Big \vert \int_0^x\big( G\circ u \big)'(t)dt  \Big \vert \leq \int_0^1  \Big\vert G'\big( u(t)  \big) u'(t) \Big\vert dt= \alpha \cdot \int_0^1  \Big\vert \big( u(t) \big)^{\alpha-1} \cdot u'(t) \Big\vert dt $
By Holder's inequality for $r$ and it's conjugate we obtain:
$\int_0^1  \Big\vert \big( u(t) \big)^{\alpha-1} \cdot u'(t) \Big\vert dt \leq \Bigg( \int_0^1 \vert u'(t)\vert^rdt \Bigg)^{\frac{1}{r}} \cdot \Bigg( \int_0^1 \vert u(t)\vert^{(\alpha-1) \frac{r}{r-1}}  dt \Bigg)^{1-\frac{1}{r}}$
Which gives us that:
$\vert u(x) \vert^\alpha \leq \alpha \Bigg( \int_0^1 \vert u'(t)\vert^rdt \Bigg)^{\frac{1}{r}} \cdot \Bigg( \int_0^1 \vert u(t)\vert^{(\alpha-1) \frac{r}{r-1}}  dt \Bigg)^{1-\frac{1}{r}}$ for all $x\in (0,1)$
Simple calculations give us that:
$1-a=\frac{q}{\alpha} \cdot (1-\frac{1}{r}) \quad$ and $\quad (\alpha-1) \cdot \frac{r}{r-1}=q$, and therefore taking the $\alpha$-th root from both sides gives us that:
$\text{(*)} \quad \vert u(x) \vert \leq \alpha^{\frac{1}{\alpha}} \Vert u' \Vert_{L^r(0,1)}^a \cdot \Vert u \Vert_{L^q(0,1)}^{(1-a)}$ for all $x\in (0,1)$. 
Remembering that $ \Vert u' \Vert_{L^r(0,1)} \leq \Vert u \Vert_{W^{1,r}(0,1)} $, this solves the case when $u(0)=0$. When $u(0)\neq 0$ using (*) for $\zeta u$ or $\zeta(x) \cdot u(1-x)$ shows that:
$\vert u(x) \vert \leq C \Vert u \Vert_{W^{1,r}(0,1)}^a \cdot \Vert u \Vert_{L^q(0,1)}^{(1-a)}$ for all $x\in (0,1)$ 
And since $u$ is continuous this shows what is needed.
