Holder's inequality $ \sum_{i=1}^n |u_i v_i| \leq (\sum_{i=1}^n |u_i|^p )^{\frac{1}{p}}(\sum_{i=1}^n |v_i|^q )^\frac{1}{q} $ Using the fact $xy \leq \frac{1}{p}x^p + \frac{1}{q}y^q$ for all $x,y >0$ and $p,q > 0$ with $\frac{1}{p} + \frac{1}{q} = 1$. How can I proof the Holder's Inequality?
$$ \sum_{i=1}^n |u_i v_i| \leq (\sum_{i=1}^n |u_i|^p )^{\frac{1}{p}}(\sum_{i=1}^n |v_i|^q )^\frac{1}{q} $$
 A: Let
$$
\begin{align}
x_i &= \frac{|u_i|}{\left(\sum_{j=1}^n |u_j|^p\right)^{1/p}}, \\
y_i &= \frac{|v_i|}{\left(\sum_{k=1}^n |v_k|^q\right)^{1/q}},
\end{align}
$$
then by $x_i y_i \le x_i^p/p + y_i^q/q$, we get
$$
\frac{|u_i|}{\left(\sum_{j=1}^n |u_j|^p\right)^{1/p}}
\frac{|v_i|}{\left(\sum_{k=1}^n |v_k|^q\right)^{1/q}}
\le
\frac{1}{p}\frac{|u_i|^p}{\sum_{j=1}^n |u_j|^p}
+
\frac{1}{q}\frac{|v_i|^q}{\sum_{k=1}^n |v_k|^q}.
$$
Summing over $i$ yields
$$
\frac{\sum_{i=1}^n|u_i||v_i|}
{\left(\sum_{j=1}^n |u_j|^p\right)^{1/p}
\left(\sum_{k=1}^n |v_k|^q\right)^{1/q}}
\le
\frac{1}{p}
+
\frac{1}{q} = 1.
$$
A: Hint:  Define $x_i$'s and $y_i$'s in terms of the $|u_i|$'s and $|v_i|$'s, but normalized in some way.  Then plug these into the inequality you mentioned, and ...
A: Define: $x_i = \dfrac{u_i}{||u||}, y_i = \dfrac{v_i}{||v||}$,and observe that : $x_1^p+\cdots +x_n^p = 1 = y_1^q+\cdots + y_n^q$, and apply the above inequality for each $x_iy_i \leq \dfrac{x_i^p}{p}+\dfrac{y_i^q}{q}$,and sum them up to get the right side being $1$.
