Why is the Cauchy Schwarz inequality a special case of Holder's inequality?

Question

--- $(1)$

--- $(2)$

Cauchy Schwarz inequality is said to be a special case of Holder's inequality when $p=2,q=2$. $(1)$ is the Holder's inequality and $(2)$ is the Cauchy Schwarz inequality. I cannot see how putting $p=2$ and $q=2$ in $(1)$ yields $(2)$. i.e. saying $|a+b|≤c$ isn't the same as saying $|a|+|b|≤c$.

Answer 1

Pugging $p=q=2$ into Holder's inequality you obtain $$ \sum_{k=1}^n\lvert a_kb_k\rvert \leq \left(\sum_{k=1}^n\lvert a_k\rvert^2\right)^{1/2}\left(\sum_{k=1}^n\lvert b_k\rvert^2\right)^{1/2} $$ Squaring either side you obtain exactly Cauchy's inequality; $$ \left(\sum_{k=1}^n\lvert a_kb_k\rvert\right)^2 \leq \left(\sum_{k=1}^n\lvert a_k\rvert^2\right)\left(\sum_{k=1}^n\lvert b_k\rvert^2\right) $$

Answer 2

Sorry for replying to an old question, but I'm reading a book where this is mentioned, and I wasn't satisfied with the other answers.

As Quoka mentions, Hölder's inequality with $p = q = 2$ yields $$ \sum_{k = 1}^n \left|a_k b_k\right| \leq \left(\sum_{k = 1}^n \left|a_k\right|^2\right)^{\frac{1}{2}} \left(\sum_{k = 1}^n \left|b_k\right|^2\right)^{\frac{1}{2}} $$

However, to obtain the Cauchy–Schwarz inequality (apparently weaker), a lower bound must be created: $$ \left|\sum_{k = 1}^n a_k b_k\right| \leq \sum_{k = 1}^n \left|a_k b_k\right| $$

This should be easy to prove, knowing that $\left|a + b\right| \leq \left|a\right| + \left|b\right|$.

It is then safe to square both sides: $$ \left|\sum_{k = 1}^n a_k b_k\right|^2 \leq \left(\sum_{k = 1}^n \left|a_k\right|^2\right) \left(\sum_{k = 1}^n \left|b_k\right|^2\right) $$

So, the Cauchy-Schwarz inequality can be seen as a special case of Hölder's inequality.