Proof of Holder's inequality for counting measure

Question

Can anyone give me a solution on how to prove Holder's inequality of this form (with the known parameters)

$$\sum_{i = 1}^n a_ib_i \le \left(\sum_{i = 1}^n a_i^p\right)^{1/p} \cdot\left(\sum_{i = 1}^n b_i^q\right)^{1/q}$$

Answer 1

It should be the following.

Let $a_i$, $b_i$, $p$ and $q$ be positives such that $\frac{1}{p}+\frac{1}{q}=1$. Prove that: $$\sum_{i = 1}^n a_ib_i \le \left(\sum_{i = 1}^n a_i^p\right)^{1/p}\left(\sum_{i = 1}^n b_i^q\right)^{1/q}.$$

Let $a_i^p=x_i$, $b_i^q=y_i$, $\frac{1}{p}=\alpha$ and $\frac{1}{q}=\beta$.

Thus, we need to prove that $$\left(\sum_{i=1}^nx_i\right)^{\alpha}\left(\sum_{i=1}^ny_i\right)^{\beta}\geq\left(\sum_{i=1}^n\left(x_i^{\alpha}y_i^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$ Indeed, let $f(x)=x^{k}$, where $k>1$.

Thus, $f$ is a convex function and by Jensen $$\sum_{i=1}^nx_it_i^k\geq\left(\sum_{i=1}^nx_it_i\right)^k,$$ for all $x_i>0$ such that $\sum\limits_{i=1}^nx_i=1$ and $t_i>0$ or $$\left(\sum_{i=1}^nx_i\right)^{k-1}\sum_{i=1}^nx_it_i^k\geq\left(\sum_{i=1}^nx_it_i\right)^k.$$ Now, take $k-1=\frac{\alpha}{\beta}$, where $\alpha>0$ and $\beta>0$, $x_it_i^k=y_i$ and we are done!