For positive finite sequence $a_{1},a_{2},...,a_{n}$ and $b_{1},b_{2},...,b_{n}$, the inequality below holds: $$ a_{1}b_{1} + ... + a_{n} b_{n} \le \left( a_{1}^{p} + ... + a_{n}^{p} \right)^{1/p} \left( b_{1}^{q} + ... + b_{n}^{q} \right)^{1/q}, \:\:\: \frac{1}{p} + \frac{1}{q} = 1 $$
Proof and attempt:
Consider the inequality:
$$ x^{\alpha} y^{\beta} \le \frac{\alpha}{\alpha + \beta} x^{\alpha + \beta} + \frac{\beta}{\alpha + \beta} y^{\alpha + \beta} $$
for all $x > 0, y > 0$, and integers $\alpha > 0$, and $\beta > 0$. This is obtained by using the general AM-GM inequality by considering $\alpha + \beta$ numbers: there are $\alpha$ $x^{\alpha + \beta}$s and $\beta$ $y^{\alpha + \beta}$s. We have by AM-GM:
$$ \frac{\alpha x^{\alpha + \beta} + \beta y^{\alpha + \beta}}{\alpha + \beta} \ge \left( x^{\alpha(\alpha + \beta)} y^{\beta(\alpha + \beta)} \right)^{\frac{1}{\alpha + \beta}} $$ $$ \frac{\alpha x^{\alpha + \beta} + \beta y^{\alpha + \beta}}{\alpha + \beta} \ge x^{\alpha} y^{\beta} $$
Now notice that if we set $ p = \frac{\alpha + \beta}{\alpha}, q = \frac{\alpha + \beta}{\beta} $ then $\frac{1}{p} + \frac{1}{q} = 1$. The inequality above becomes:
$$ x^{\alpha} y^{\beta} \le \frac{1}{p} x^{\alpha p} + \frac{1}{q} y^{\beta q} $$
Now let $u = x^{\alpha}, v = y^{\beta}$:
$$ u v \le \frac{1}{p} u^{p} + \frac{1}{q} v^{q} $$
Now use this many times: $u = a_{i}, v = b_{i}, \:\: i = 1, ..., n$, and then sum all of the $n$ inequalities to get:
$$ a_{1}b_{1} + ... + a_{n} b_{n} \le \frac{1}{p} \sum_{i=1}^{n} a_{i}^{p} + \frac{1}{q} \sum_{i=1}^{n} b_{i}^{q} $$
Now we use the fact that $\frac{1}{p} + \frac{1}{q}=1$, by the transformation $a_{i} = \frac{a_{i}^{*}}{ \left( \sum_{k=1}^{n} (a_{k}^{*})^{p} \right)^{1/p} }$ and $b_{i} = \frac{b_{i}^{*}}{ \left( \sum_{k=1}^{n} (b_{k}^{*})^{q} \right)^{1/q} }$.
$$ \frac{ \sum_{k=1}^{n} a_{k}^{*} b_{k}^{*} }{ \left( \sum_{k=1}^{n} (a_{k}^{*})^{p} \right)^{1/p} \left( \sum_{k=1}^{n} (b_{k}^{*})^{q} \right)^{1/q} } $$ $$ \le \frac{1}{p} \sum_{i=1}^{n} \frac{(a_{i}^{*})^{p}}{ \sum_{k=1}^{n} (a_{k}^{*})^{p} } + \frac{1}{q} \sum_{i=1}^{n} \frac{ (b_{i}^{*})^{q}}{ \sum_{k=1}^{n} (b_{k}^{*})^{q} } $$
$$ \frac{ \sum_{k=1}^{n} a_{k}^{*} b_{k}^{*} }{ \left( \sum_{k=1}^{n} (a_{k}^{*})^{p} \right)^{1/p} \left( \sum_{k=1}^{n} (b_{k}^{*})^{q} \right)^{1/q} } $$ $$ \le \frac{1}{p} + \frac{1}{q} = 1 $$
$$ \sum_{k=1}^{n} a_{k}^{*} b_{k}^{*} \le \left( \sum_{k=1}^{n} (a_{k}^{*})^{p} \right)^{1/p} \left( \sum_{k=1}^{n} (b_{k}^{*})^{q} \right)^{1/q} $$
Now for the equality, here is my attempt..which is a bit inconsistent with the one from The Cauchy-Schwarz Master Class . Therefore I would like verification whether or not the author meant something different, or maybe typo.
Using our last inequality of our proof, equality is when:
$$ \sum_{k=1}^{n} a_{k}^{*} b_{k}^{*} = \left( \sum_{k=1}^{n} (a_{k}^{*})^{p} \right)^{1/p} \left( \sum_{k=1}^{n} (b_{k}^{*})^{q} \right)^{1/q} $$ If we recall the form $a_{k}, b_{k}$ to a form involving $a_{k}^{*}, b_{k}^{*}$, the above is equivalent to
$$ \sum_{k=1}^{n} a_{k} b_{k} = \frac{1}{p} \sum_{i=1}^{n} a_{i}^{p} + \frac{1}{q} \sum_{i=1}^{n} b_{i}^{q} $$
which is also equivalent with
$$ a_{k} b_{k} = \frac{a_{k}^{p}}{p} + \frac{b_{k}^{q}}{q}, \:\: k=1,2,...,n $$
recall that this is the result from our general AM-GM usage in the proof, when equality happens. That is when $a_{k}^{p} = b_{k}^{q}$. Using our previous transformation again, the equation is equivalent with
$$ \left( \frac{a_{k}^{*}}{ \left( \sum_{i=1}^{n} (a_{i}^{*})^{p} \right)^{1/p} } \right)^{p} = \left( \frac{b_{k}^{*}}{ \left( \sum_{i=1}^{n} (b_{i}^{*})^{q} \right)^{1/q} } \right)^{q} $$
$$ \frac{(a_{k}^{*})^{p}}{ \left( \sum_{i=1}^{n} (a_{i}^{*})^{p} \right) } = \frac{(b_{k}^{*})^{q}}{ \left( \sum_{i=1}^{n} (b_{i}^{*})^{q} \right) } $$
$$ \frac{\left( \sum_{i=1}^{n} (b_{i}^{*})^{q} \right) (a_{k}^{*})^{p}}{ \left( \sum_{i=1}^{n} (a_{i}^{*})^{p} \right) } = (b_{k}^{*})^{q} $$
Write $\lambda = \frac{\left( \sum_{i=1}^{n} (b_{i}^{*})^{q} \right)}{ \left( \sum_{i=1}^{n} (a_{i}^{*})^{p} \right) } $ and we get the equality condition $ \lambda (a_{k}^{*})^{p} = (b_{k}^{*})^{q} $ (this is a bit different with the result from the book)