Prove Holder inequality by using $a^{\theta}b^{1-\theta}\leq\theta a+(1-\theta)b$ for all $a,b\geq0$ and $\theta\in[0,1]\\$

Question

Known that $a^{\theta}b^{1-\theta}\leq\theta a+(1-\theta)b$ for all $a,b\geq0$ and $\theta\in[0,1]\\$. Now I want to prove, for $x,y\in R^n$ and $p,q>1$ s.t. $\frac{1}{p}+\frac{1}{q}=1$: $\sum_{i=1}^{n}{x_iy_i} \leq (\sum_{i=1}^{n}|x_i|^p)^{\frac{1}{p}}(\sum_{i=1}^{n}|y_i|^q)^{\frac{1}{q}}$. Here is the hint: $let\quad a = \frac{|x_i|^p}{\sum_{j=1}^{n}|x_j|^p}\quad b = \frac{|y_i|^q}{\sum_{j=1}^{n}|y_j|^q}$ and $\theta=\frac{1}{p}.$ $\quad$ I've tried to plug in the formula as hint indicated and moved the RHS of the desired inequality to one side but don't know what to do next.

Answer 1

Assume first that $\sum |x_i|^{p}=\sum |y_i|^{q}=1$. Then we get $|x_i|^{p\theta}|y_i|^{q(1-\theta)}=|x_iy_i| $. Hence $\sum |x_iy_i| \leq \sum _i(\frac 1 p |x_i|^{p} +\frac 1 q |y_i|^{q}) \leq 1$. This proves the result when $\sum |x_i|^{p}=\sum |y_i|^{q}=1$. For the general case just apply what you have proved with $x_i$ replaced by $\frac {x_i} {(\sum |x_i|^{p})^{1/p}}$ and $y_i$ replaced by $\frac {y_i} {(\sum |q_i|^{p})^{1/q}}$.

Proof without going to the special case:

Let $A=(\sum |x_i|^{p})^{1/p}$ and $A=(\sum |x_i|^{p})^{1/p}$. Then $\frac 1 A|x_i| \frac 1 B|y_i| \leq \frac 1 p |x_i|^{p}/A^{p}+\frac 1 q |y_i|^{q}/A^{q}$. Summing this and multiplying by $AB$ we get $\sum |x_iy_i| \leq \frac 1 p(\sum |x_i|^{p} /A^{p-1}+\frac 1 q(\sum |y_i|^{q} /B^{q-1}$. Now can you finish the proof?