# Proving Holder's Inequality from $a^nb^{(1-n)} \leq na + (1-n)b$

$a^nb^{(1-n)} \leq na + (1-n)b$, $a,b \geq 0$, $0\leq n\leq 1$

Let $a = \dfrac{|x_i|^p}{\sum\limits_{j=1}^n |x_j|^p}$, $b = \dfrac{|y_i|^q}{\sum\limits_{j=1}^n |y_j|^q}$, n =$1/p$

Then plugging everything in we have:

$(\dfrac{|x_i|^p}{\sum\limits_{j=1}^n |x_j|^p})^{1/p} (\dfrac{|y_i|^q}{\sum\limits_{j=1}^n |y_j|^q})^{1/q} \leq \dfrac{|x_i|^p}{p\sum\limits_{j=1}^n |x_j|^p} + \dfrac{|y_i|^q}{q\sum\limits_{j=1}^n |y_j|^q}$

So if we multiply by the denominator on the left hand side (i.e. $\|x\|_p\|y\|_q$) on both sides:

$|x_i||y_i| \leq \frac{1}{p}|x_i|^p \|x\|^{-(p-1)}_p \|y\|_q + \frac{1}{q}|y_i|^q\|x\|_p\|y\|^{-(q-1)}_q$

I am not sure how to proceed from here, can someone click click?

Starting with the last line of your proof, observe that: $\displaystyle \sum_{k=1}^n |x_i|^p = (||x||_p)^p, \displaystyle \sum_{k=1}^n |y_i|^q = (||y||_q)^q$, and add these inequalities up, your right hand side is: $\left(\dfrac{1}{p}+\dfrac{1}{q}\right)||x||_p||y||_q = ||x||_p||y||_q$, and you are done.
• Just to clarify, so you put summation sign on the last line i.e. $\sum_{i = 1}^n |x_i||y_i| \leq \sum_{i = 1}^n \frac{1}{p}|x_i|^p \|x\|^{-(p-1)}_p \|y\|_q + \sum_{i = 1}^n \frac{1}{q}|y_i|^q\|x\|_p\|y\|^{-(q-1)}_q$ but you only summed over the $|x_i|^p$ and $|y_i|^q$ terms is it because the terms in $\| \|$ are already scalars so you can just pull them out? Good answer by the way Commented Oct 13, 2015 at 21:35