Cauchy-Schwarz by induction

Base case $n=2$. $(a_1b_1+a_2b_2)^2\le(a_1^2+a_2^2)(b_1^2+b_2^2)$ (proved earlier).

Inductive step. Assume the inequality for $n-1$ terms. Let $S_n=\sum_{i=1}^n a_ib_i$. By Cauchy-Schwarz on the last two "blocks" $(\sum_{i<n}a_ib_i)$ and $(a_nb_n)$:
$$S_n^2=(S_{n-1}+a_nb_n)^2\le(S_{n-1}^2+a_n^2)(1+b_n^2/a_n^2)?$$
This is not the standard approach.

Correct approach. By the inductive hypothesis,
$$S_{n-1}^2\le\left(\sum_{i<n}a_i^2\right)\left(\sum_{i<n}b_i^2\right).$$
Then apply the $2$-variable inequality to $S_{n-1}$ and $a_nb_n$:
$$S_n^2\le\left(\sqrt{\sum_{i<n}a_i^2}\sqrt{\sum_{i<n}b_i^2}+a_nb_n\right)^2.$$
This needs to be bounded by $(\sum_{i\le n}a_i^2)(\sum_{i\le n}b_i^2)$, which follows after Cauchy-Schwarz on the last step.

Conclusion. Cauchy-Schwarz holds for all $n$.