Pseudo Otto Holder proof help.

I'm at the first part of a pseudo Otto Holder theorem which claims if $1/p + 1/q = 1$ where $x \in \ell_p$ and $a \in \ell_q$ then $\sum_\limits{i=0}^\infty |a_i x_i| \leq ||a||_q||x||_p$.

For the first part we're allowed to use the fact that for any positive numbers $\alpha, \beta \in \mathbb{R}_{+}$ and any $\lambda \in [0,1]$, we have $\alpha^\lambda \beta^{1-\lambda} \leq \lambda \alpha + (1-\lambda) \beta$. To show that for any $i \in \mathbb{N}$,

$$ \cfrac{|a_i x_i|}{||x||_p||a||_q} \leq \cfrac{1}{p} \left(\cfrac{|x_i|}{||x||_p}\right)+\cfrac{1}{q} \left(\cfrac{|a_i|}{||a||_q}\right) $$

Let $1/p = \lambda \implies 1 - \lambda = 1/q$

$$ \cfrac{|a_i x_i|}{||x||_p||a||_q} \leq \lambda \left(\cfrac{|x_i|}{||x||_p}\right)+(1-\lambda) \left(\cfrac{|a_i|}{||a||_q}\right) $$

I can tell that this almost looks like the homework fact we're told to use but I'm not sure how to get some lambda in the powers on the left hand side... Some diction would be nice.

Update

$$ \begin{align*} \cfrac{|a_i x_i|}{||x||_p ||a||_q} &= \cfrac{|x_i|}{||x||_p} \cfrac{|a_i|}{||a||_q} \\ &= \left(\left(\cfrac{|x_i|}{||x||_p}\right)^{\lambda}\right)^{1/\lambda} \left(\left(\cfrac{|a_i|}{||a||_q}\right)^{1-\lambda}\right)^{1/(1-\lambda)}\\ &\leq \lambda \left(\cfrac{|x_i|}{||x||_p}\right)+(1-\lambda)\left(\cfrac{|a_i|}{||a||_q}\right) \\ \text{Set $\lambda = 1/p$ which implies $1 - \lambda = 1/q$}& \\ &= \cfrac{1}{p} \left(\cfrac{|x_i|}{||x||_p}\right) + \cfrac{1}{q}\left(\cfrac{|a_i|}{||a||_q}\right) \\ \end{align*} $$

Which is what I was after. Pretty easy once you know the trick! Thanks @PatrickR

Image of the homework problem.

Assume $||a||_q$ and $||x||_p$ are not zero. To show $\sum |a_i x_i| \leq ||a||_p||x||_q$, it is equivalent to show $$\sum(\cfrac{|a_i|}{||a||_q})(\cfrac{|x_i|}{||x||_p}) \le 1\;.$$ So by dividing the two vectors by their norms you can assume without loss of generality that $||a||_q=1$ and $||x||_p=1$, and you have to show $\sum|a_ix_i|\le 1$.

Now to make use of the inequality you want to use, the trick is to rewrite $$|a_ix_i|=(|a_i|^q)^{1/q}(|x_i|^p)^{1/p}$$ which is of the form $\alpha^\lambda \beta^{1-\lambda}$ with $\lambda=1/q$ and $1-\lambda=1/p$.

The rest should follow easily.