Given that $a_i, b_i > 0$ and that $ p, q > 1$ and $\frac{1}{p} + \frac{1}{q} = 1$ I want to show Holder's inequality, that $\sum_{i}a_ib_i \leq (\sum_{i} a_i ^p)^{\frac{1}{p}} (\sum_{i} b_i ^q)^{\frac{1}{q}}$ and that equality is attained when $\frac{a_i}{b_i}$ = constant.
We are given the hint: Assume WLOG that $\sum_{i}a_i^p=\sum_{i}b_i^q = 1$
What I did:
Starting from $\sum_{i}a_ib_i$, use Young's inequality with the given $p, q$:
$\sum_{i}a_ib_i \leq\sum_{i}[\frac{a_i ^p}{p} + \frac{b_i^q}{q}] = \frac{1}{p}\sum_i a_i ^p + \frac{1}{q}\sum b_i^q$
Now use our WLOG assumption to get $\sum_{i}a_ib_i \leq \frac{1}{p} + \frac{1}{q} = 1 = 1 \cdot1=1^{\frac{1}{p}} \cdot 1^{\frac{1}{q}} =(\sum_{i} a_i ^p)^{\frac{1}{p}} (\sum_{i} b_i ^q)^{\frac{1}{q}} $
My problem is that this seems overly specific. I didn't really prove Holder's inequality, only for this specific case. Maybe it's just a feeling but the assumption feels too powerful and I don't see how to generalize it easily.
Is this proof correct? It feels off.
Let $c_{i}=\dfrac{a_{i}}{\|a\|_{p}}$, $d_{i}=\dfrac{b_{i}}{\|b\|_{q}}$, where $a=(a_{i})$, $b=(b_{i})$, then $c:=(c_{i})$ is such that $\|c\|_{p}=1$, similar to $\|d\|_{q}=1$, so $\displaystyle\sum_{i}c_{i}d_{i}\leq 1$ and hence $\displaystyle\sum_{i}a_{i}b_{i}\leq\|a\|_{p}\|b\|_{q}$.