Proving Holder's inequality for sums I want to prove the Holder's inequality for sums:

Let $p\ge1$
    be a real number. Let $(x_{k})\in l_{p}$
    and $(y_{k})\in l_{q}$
   . Then,
  $$\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}y_{k}\vert\le\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}$$
  with $q\in\mathbb{R}$
    such that $\frac{1}{p}+\frac{1}{q}=1$
   .

Inspired by a proof I've seen before, I attempted at a solution. I would like you to confirm the ideas and to answer the questions, which correspond to steps of the proof I don't know how to justify.
My attempt: Let $p>1$
  be a real number. $Let (x_{k})\in l_{p}$
  and $(y_{k})\in l_{q}$
 .
If $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}=0$
  or $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}=0$
  the inequality is trivially true (Question1: Is it? Why?)
In case both are nonzero, we can define the sequences $(z_{k})$
  and $(w_{k})$
  with
$$z_{k}=\frac{x_{k}}{\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}}\ \text{ and }\ w_{k}=\frac{y_{k}}{\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}}$$
Now, $$\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)$$
  (by Young's Inequality).
But $$\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)=\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert x_{k}\vert^{p}}{p\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)}+\frac{\vert y_{k}\vert^{q}}{q\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)}\right)\le1$$
 (Question2: Is this last inequality true? Why? If it is, the result follows smoothly...)
So $\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le1$
 . Multiplying both sides by the (positive) term $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\cdot\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}$
 , it's done. $\square$
 A: For completeness I'll leave here my complete proof, using the suggestions in the comments.
Proof: Let $p>1$
  be a real number. $Let (x_{k})\in l_{p}$
  and $(y_{k})\in l_{q}$
 .
If $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}=0$
  or $\left(\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}\right)^{\frac{1}{q}}=0$
  the inequality is true: This is equivalent to $\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}=0$
  or $\overset{\infty}{\underset{k=1}{\sum}}\vert y_{k}\vert^{q}=0$
 , and, if any term of one of these series is nonzero, the series would be nonzero (because all terms are zero or positive), so $\vert x_{k}\vert^{p}=0$
  or $\vert y_{k}\vert^{q}=0$
  for all $k$
 . Hence, $\vert x_{k}\vert^{p}\vert y_{k}\vert^{q}=0$
  for all $k$.
In case both are nonzero, we can define the sequences $(z_{k})$
  and $(w_{k})$
  with
$$z_{k}=\frac{x_{k}}{\left(\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert^{p}\right)^{\frac{1}{p}}}\ \text{ and }\ w_{k}=\frac{y_{k}}{\left(\overset{\infty}{\underset{l=1}{\sum}}\vert y_{l}\vert^{q}\right)^{\frac{1}{q}}}.$$
Now, $$\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)$$
  by Young's Inequality.
But $$\overset{\infty}{\underset{l=1}{\sum}}\vert z_{l}\vert^{p}=\overset{\infty}{\underset{l=1}{\sum}}\frac{\vert x_{l}\vert^p}{\Big\vert\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\Big\vert^p}=\frac{\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert^p}{\Big\vert\left(\overset{\infty}{\underset{k=1}{\sum}}\vert x_{k}\vert^{p}\right)^{\frac{1}{p}}\Big\vert^p}=1$$
  and similarly $$\overset{\infty}{\underset{l=1}{\sum}}\vert w_{l}\vert^{q}=1.$$
Hence $$\overset{\infty}{\underset{k=1}{\sum}}\left(\frac{\vert z_{k}\vert^{p}}{p}+\frac{\vert w_{k}\vert^{q}}{q}\right)=\frac{1}{p}+\frac{1}{q}=1.$$
So $\overset{\infty}{\underset{k=1}{\sum}}\vert z_{k}w_{k}\vert\le1$
 . Multiplying both sides by the (positive) term $\left(\overset{\infty}{\underset{l=1}{\sum}}\vert x_{l}\vert^{p}\right)^{\frac{1}{p}}\cdot\left(\overset{\infty}{\underset{l=1}{\sum}}\vert y_{l}\vert^{q}\right)^{\frac{1}{q}}$
 , it's done. $\square$
